Find the rth term of the AP, the sum of whose first n terms is
Question: Find the $r^{\text {th }}$ term of the AP, the sum of whose first $n$ terms is $\left(3 n^{2}+2 n\right)$. Solution: Given: The sum of first n terms. To Find: The $\mathrm{r}^{\text {th }}$ term. Let the first term be a and common difference be d Put n = 1 to get the first term $a=S_{1}=3+2=5$ Put $n=2$ to get $a+(a+d) 2 a+d=12+4=1610+d=16 d=6 t_{r}=a+(r-1) d$ $t \gg r=5+(r-1) 6=5+6 r-6=6 r-1$ The $\mathrm{r}^{\text {th }}$ term is given by $6 \mathrm{r}-1$....
Read More →Find the value of x such that
Question: Find the value of x such that 25 + 22 + 19 + 16 + . + x = 112. Solution: To Find: The value of x, i.e. the last term. Given: The series and its sum. The series can be written as x, (x + 3), , 16, 19, 22, 25 Let there be n terms in the series $25=x+(n-1) 3$ $3(n-1)=25-x x=25-3(n-1)=28-3 n$ Let S be the sum of the series $S=\frac{n}{2}[x+25]=112$ $\Rightarrow \mathrm{n}[28-3 \mathrm{n}+25]=224$ $\Rightarrow \mathrm{n}(53-3 \mathrm{n})=224$ $\Rightarrow 3 \mathrm{n}^{2}-53 \mathrm{n}+224=...
Read More →Find the sum of the series
Question: Find the sum of the series 1 + 4 + 7 + 10 + . + x = 715. Solution: Note: The sum of the series is already provided in the question. The solution to find x is given below. Let there be n terms in the series $x=1+(n-1) 3$ $=3 n-2$ Let S be the sum of the series $S=\frac{n}{2}[1+x]=715$ $\Rightarrow \mathrm{n}[1+3 \mathrm{n}-2]=1430$ $\Rightarrow \mathrm{n}+3 \mathrm{n}^{2}-2 \mathrm{n}=1430$ $\Rightarrow 3 \mathrm{n}^{2}-\mathrm{n}-1430=0$ Applying Sri Dhar Acharya formula, we get $\math...
Read More →Find the sum of the series
Question: Find the sum of the series 101 + 99 + 97 + 95 + . + 43. Solution: To Find: The sum of the given series. Sum of the series is given by $S=\frac{n}{2}(a+l)$ Where n is the number of terms, a is the first term and l is the last term Here $a=101, I=43, n=30$ $S=\frac{30}{2}[101+43]$ = 15 144 = 2160 The sum of the series is 2160....
Read More →Find the sum of the series
Question: Find the sum of the series 2 + 5 + 8 + 11 + . + 191. Solution: To Find: The sum of the given series. The nth term of an AP series is given by $t_{n}=a+(n-1) d$ $\Rightarrow 191=2+(n-1) 3$ $\Rightarrow 3(n-1)=189$ $\Rightarrow n-1=63$ $\Rightarrow n=64$ Therefore, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$ $S_{n}=\frac{64}{2}[4+63 \times 3]$ = 32 193 = 6176 The sum of the series is 6176....
Read More →Find the sum of n term of the A P
Question: Find the sum of $n$ term of the $A P \frac{x-y}{x+y}, \frac{3 x-2 y}{x+y}, \frac{5 x-3 y}{x+y}, \ldots .$ Solution: To Find: The sum of n terms of the given AP. Sum of n terms of an AP with first term a and common difference d is given by $S=\frac{n}{2}[2 a+(n-1) d]$ Here $a=x-y, d=2 x-y$ $\Rightarrow S=\frac{1}{x+y} \times \frac{n}{2} \times[2 x-2 y+(n-1)(2 x-y)]$ $\Rightarrow S=\frac{n}{2(x+y)}[2 x-2 y+n(2 x-y)-2 x+y]$ $\Rightarrow S=\frac{n}{2(x+y)}[n(2 x-y)-y]$ The sum of the serie...
Read More →Find the sum of 20 terms of the AP
Question: Find the sum of 20 terms of the AP $(x+y),(x-y),(x-3 y), \ldots$ Solution: To Find: The sum of 20 terms of the given AP. Sum of n terms of an AP with first term a and common difference d is given by $S=\frac{n}{2}[2 a+(n-1) d]$ Here a = x + y, n = 20, d = - 2y $\Rightarrow S=10[2 x+2 y+19(-2 y)]=10[2 x+2 y-38 y]=10[2 x-36 y]$ $\Rightarrow S=20[x-18 y]$ Sum of the series is 20(x - 18y)....
Read More →Find the sum of 100 term of the AP
Question: Find the sum of 100 term of the AP $0.6,0.61,0.62,0.63, \ldots$ Solution: To Find: The sum of 100 terms of the given AP series. Sum of n terms of an AP with first term a and common difference d is given by $S=\frac{n}{2}[2 a+(n-1) d]$ Here a = 0.6, n = 100, d = 0.01 $\Rightarrow S=\frac{100}{2}[1.2+99 \times 0.01]$ $=50[1.2+0.99]$ $=50 \times 2.19$ $109.5$ Sum of the series is 109.5...
Read More →Find the sum of 25 terms of the AP
Question: Find the sum of 25 terms of the AP $\sqrt{2}, 2 \sqrt{2}, 3 \sqrt{2}, 4 \sqrt{2}, \ldots$ Solution: To Find: The sum of 25 terms of the given AP series Sum of $n$ terms of an $A P$ with first term a and common difference $d$ is given by $S=\frac{n}{2}[2 a+(n-1) d]$ Here, $a=\sqrt{2}, n=25, d=\sqrt{2} \Rightarrow S=\frac{25}{2}[2 \sqrt{2}+24 \sqrt{2}]$ $=25 \times 13 \times \sqrt{2}=325 \sqrt{2}$ Sum of 25 terms is $325 \sqrt{2}$....
Read More →Find the sum of 16 terms of the
Question: Find the sum of 16 terms of the AP $6,5 \frac{1}{3}, 4 \frac{2}{3}, 4, \ldots$ Solution: To find: Sum of 16 terms of the AP Given: First term = 6 Common difference $=-\frac{2}{3}$ $\Rightarrow S_{n}=\frac{n}{2}[2 a+(n-1) d]$ $\Rightarrow S_{n}=\frac{16}{2}\left[2 \times 6+15 \times\left(-\frac{2}{3}\right)\right]$ $\Rightarrow S_{n}=\frac{16}{2}[12-10]_{S_{n}}=16$ The sum of first 16 terms of the series is 16...
Read More →Find the sum of 23 terms of the AP
Question: Find the sum of 23 terms of the AP $17,12,7,2,-3, \ldots$ Solution: To Find: The sum of 25 terms of the given AP series. Sum of n terms of an AP with first term a and common difference d is given by $S=\frac{n}{2}[2 a+(n-1) d]$ Here, a = 17, n = 23 and d = - 5 $S=\frac{23}{2}[34+22(-5)]$ $\Rightarrow S=\frac{23}{2}[34-110]=\frac{23}{2} \times(-76)$ $=-874$ Sum of 23 terms of the AP IS 874....
Read More →A man starts repaying a loan as the first instalment of 10000.
Question: A man starts repaying a loan as the first instalment of 10000. If he increases the instalment by 500 every month, what amount will he pay in 30thinstalment? Solution: To Find: what amount will he pay in the 30th instalment. Given: first instalment =10000 and it increases the instalment by 500 every month. $\therefore$ So it form an AP with first term is 10000 , common difference 500 and number of instalment is 30 Formula Used: $T_{n}=a+(n-1) d$ (Where $\mathrm{a}$ is first term, $\math...
Read More →A side of an equilateral triangle is 24 cm long.
Question: A side of an equilateral triangle is 24 cm long. A second equilateral triangle is inscribed in it by joining the midpoints of the sides of the first triangle; the process is continued. Find the perimeter of the sixth inscribed equilateral triangle. Solution: To Find: The perimeter of the sixth inscribed equilateral triangle. 1st Given: Side of an equilateral triangle is $24 \mathrm{~cm}$ long. As 2nd triangle is formed by joining the midpoints of the sides of the first triangle whose s...
Read More →We know that the sum of the interior angles of a triangle is 180°.
Question: We know that the sum of the interior angles of a triangle is 180. Show that the sum of the interior angles of polygons with 3, 4, 5, 6, . sides form an arithmetic progression. Find the sum of the interior angles for a 21 - sided polygon. Solution: Show that: the sum of the interior angles of polygons with 3, 4, 5, 6, . sides form an arithmetic progression. To Find: The sum of the interior angles for a 21 - sided polygon. Given: That the sum of the interior angles of a triangle is 180. ...
Read More →Find the number of terms common to the two arithmetic progressions 5, 9,
Question: Find the number of terms common to the two arithmetic progressions 5, 9, 13, 17, ., 217 and 3, 9, 15, 21, ., 321. Solution: To Find: The number of terms common to both AP Given: The 2 APs are 5, 9, 13, 17, ., 217 and 3, 9, 15, 21, ., 321 As we find that first common term of both AP is 9 and the second common term of both AP is 21 Let suppose the new AP whose first term is 9, the second term is 21, and the common difference is 21 9 = 12 NOTE: As first AP the last term is 217 and second ...
Read More →The digits of a 3 - digit number are in AP, and their sum is 15.
Question: The digits of a 3 - digit number are in AP, and their sum is 15. The number obtained by reversing the digits is 594 less than the original number. Find the number. Solution: To Find: The number Given: The digits of a 3 digit number are in AP, and their sum is 15. Let required digit of 3 - digit number be (a - d), (a), (a + d). Then, $(a-d)+(a)+(a+d)=15 \Rightarrow 3 a=15 \Rightarrow a=5$ (Figure show 3 digit number original number) (Figure show 3 digit number in reversing form) So, $(5...
Read More →The angles of a quadrilateral are in AP
Question: The angles of a quadrilateral are in AP whose common difference is 10. Find the angles. Solution: To Find: The angles of a quadrilateral. Given: Angles of a quadrilateral are in AP with common difference $=10^{\circ}$. Let the required angles be $a,\left(a+10^{\circ}\right),\left(a+20^{\circ}\right)$ and $\left(a+30^{\circ}\right)$ Then, $a+\left(a+10^{\circ}\right)+\left(a+20^{\circ}\right)+\left(a+30^{\circ}\right)=360^{\circ} \Rightarrow 4 a+60^{\circ}=360^{\circ} \Rightarrow a=75^{...
Read More →The sum of three consecutive terms of an AP
Question: The sum of three consecutive terms of an AP is 21, and the sum of the squares of these terms is 165. Find these terms Solution: To Find: The three numbers which are in AP. Given: Sum and sum of the squares of three numbers are 21 and 165 respectively. Let required number be $(a-d),(a),(a+d) .$ Then, $(a-d)+a+(a+d)=21 \Rightarrow 3 a=21 \Rightarrow a=7$ Thus, the numbers are $(7-d), 7$ and $(7+d)$. But their sum of the squares of three numbers is 165. $\therefore(7-\mathrm{d})^{2}+7^{2}...
Read More →Three numbers are in AP. If their sum is 27 and their product is 648,
Question: Three numbers are in AP. If their sum is 27 and their product is 648, find the numbers. Solution: To Find: The three numbers which are in AP. Given: Sum and product of three numbers are 27 and 648 respectively. Let required number be $(a-d),(a),(a+d)$. Then, $(a-d)+a+(a+d)=27 \Rightarrow 3 a=27 \Rightarrow a=9$ Thus, the numbers are $(9-d), 9$ and $(9+d)$ But their product is 648 . $\therefore(9-d) \times 9 \times(9+d)=648$ $\Rightarrow(9-d)(9+d)=72$ $\Rightarrow 81-d^{2}=72 \Rightarro...
Read More →In an AP, it is being given that
Question: In an AP, it is being given that $\frac{T_{4}}{T_{7}}=\frac{2}{3} .$ Find $\frac{T_{7}}{T_{10}}$ Solution: To Find: $\frac{T 7}{T 10}$ Given: $\frac{T_{4}}{T_{7}}=\frac{2}{3}$ (Where $T_{n}$ is nth term and $d$ is common difference of given AP) Formula Used: $T_{n}=a+(n-1) d$ $\frac{\mathrm{T}_{4}}{\mathrm{~T}_{7}}=\frac{2}{3} \rightarrow \frac{a+3 d}{a+6 d}=\frac{2}{3}$ (cross multiply) $3 a+9 d=2 a+12 d \rightarrow a=3 d$.......equation (i) Now $\frac{\mathrm{T}_{7}}{\mathrm{~T}_{10}...
Read More →Solve this
Question: If $\theta_{1}, \theta_{2}, \theta_{3}, \ldots, \theta_{n}$ are in AP whose common difference is $d$, show that $\sec \theta_{1} \sec \theta_{2}+\sec \theta_{2} \sec \theta_{3}+\ldots . .+\sec \theta_{n-1} \sec \theta n=\frac{\left(\tan \theta_{n}-\tan \theta_{1}\right)}{\sin d}$ Solution: Show that: $\sec \theta_{1} \sec \theta_{2}+\sec \theta_{2} \sec \theta_{3}+\ldots .+\sec \theta_{n-1} \sec \theta n=\frac{\left(\tan \theta_{n}-\tan \theta_{1}\right)}{\sin d}$ Given: Given AP is $\...
Read More →How many 2 - digit numbers are divisible by 3?
Question: How many 2 - digit numbers are divisible by 3? Solution: To Find : 2 - digit numbers divisible by 3. First 2 - digit number divisible by 3 is 12 Second 2 - digit number divisible by 3 is 15 and Last 2 - digit number divisible by is 99 . Given: The AP is $12,15,18, \ldots \ldots \ldots \ldots, 99$ $a_{1}=12, a_{2}=15, d=15-12=3$ and $a_{n}=99$ (Where $a=a_{1}$ is First term, $a_{2}$ is Second term, $a_{n}$ is nth term and $d$ is common difference of given $\mathrm{AP}$ ) Formula Used: $...
Read More →How many 3 - digit numbers are divisible by 7?
Question: How many 3 - digit numbers are divisible by 7? Solution: To Find : 3 - digit numbers divisible by 7. First 3 - digit number divisible by 7 is 105 Second 3 - digit number divisible by 7 is 112 and Last 3 - digit number divisible by 7 is 994 . Given: The AP is $105,112,119$, , 994 $a_{1}=105, a_{2}=112, d=112-105=7$ and $a_{n}=994$ (Where $a=a_{1}$ is First term, $a_{2}$ is Second term, $a_{n}$ is nth term and $d$ is common difference of given $\mathrm{AP}$ ) Formula Used: $a_{n}=a+(n-1)...
Read More →Find the 16th term from the end of the AP
Question: Find the $16^{\text {th }}$ term from the end of the AP $7,2,-3,-8,-13, \ldots,-113$ Solution: To Find : 28th term from the end of the AP. Given: The AP is $7,2,-3,-8,-13, \ldots,-113$ $a_{1}=7, a_{2}=2, d=2-7=-5$ and $I=-113$ Formula Used: nth term from the end $=1-(n-1) d$ (Where $/$ is last term and $d$ is common difference of given $A P$ ) By using $n$th term from the end $=1-(n-1) d$ formula 16th term from the end $=(-113)-15 d \rightarrow(-113)-15 \times(-5)=-38$ So $16^{\text {t...
Read More →Find the 28th term from the end of the AP
Question: Find the $28^{\text {th }}$ term from the end of the AP $6,9,12,15,18, \ldots ., 102 .$ Solution: To Find $28^{\text {th }}$ term from the end of the AP. Given: The AP is $6,9,12,15,18, \ldots ., 102$ $a_{1}=6, a_{2}=9, d=9-6=3$ and $I=102$ Formula Used: $n$th term from the end $=1-(n-1) d$ (Where lis last term and $d$ is common difference of given AP) By using $n$th term from the end $=1-(n-1) d$ formula 28th term from the end $=102-27 \mathrm{~d} \rightarrow 102-27^{\times} 3=21$ So ...
Read More →