Solve this
Question: If $S_{m}=m^{2} p$ and $S_{n}=n^{2} p$, where $m \neq n$ in an AP then prove that $S_{p}=p^{3}$. Solution: Let the first term of the AP be a and the common difference be d Given: $S_{m}=m^{2} p$ and $S_{n}=n^{2} p$ To prove: $\mathrm{Sp}=\mathrm{p}^{3}$ According to the problem $\frac{m}{2}[2 a+(m-1) d]=m^{2} p \Rightarrow 2 a+(m-1) d=2 m p$ and $\frac{n}{2}[2 a+(n-1) d]=n^{2} p \Rightarrow 2 a+(n-1) d=2 n p$ Subtracting the equations we get, $(m-n) d=2 p(m-n)$ Now m is not equal to n ...
Read More →How many non-overlapping triangles
Question: How many non-overlapping triangles can we make in a-n-gon (polygon having n sides), by joining the vertices? (a)n-1 (b)n-2 (c) n 3 (d) n 4 Solution: (b) The number of non-overlapping triangles in a n-gon = n 2, i.e. 2 less than the number of sides....
Read More →What is the maximum number of obtuse
Question: What is the maximum number of obtuse angles that a quadrilateral can have? (a) 1 (b) 2 (c) 3 (d) 4 Solution: (c) We know that, the sum of all the angles of a quadrilateral is 360. Also, an obtuse angle is more than 90 and less than 180. Thus, all the angles of a quadrilateral cannot be obtuse. Hence, almost 3 angles can be obtuse....
Read More →Which of the following is a propefay
Question: Which of the following is a propefay of a parallelogram? (a) Opposite sides are parallel (b) The diagonals bisect each other at right angles (c) The diagonals are perpendicular to each other (d) All angles are equal Solution: (a) We,know that, in a parallelogram, opposite sides are parallel....
Read More →Which of the following properties
Question: Which of the following properties describe a trapezium? (a) A pair of opposite sides is parallel (b) The diagonals bisect each other (c) The diagonals are perpendicular to each other (d) The diagonals are equal Solution: (a) We know that, in a trapezium, a pair of opposite sides are parallel....
Read More →For which of the following figures,
Question: For which of the following figures, diagonals are equal? (a) Trapezium (b) Rhombus (c) Parallelogram (d) Rectangle Solution: (d) By the property of a rectangle, we know that its diagonals are equal....
Read More →If the sum of n terms of an AP is
Question: If the sum of n terms of an AP is $\left\{n P+\frac{1}{2} n(n-1) Q\right\}$ where P and Q are constants then find the common difference. Solution: Let the first term be a and common difference be d To Find: d Given: Sum of $n$ terms of $A P=n P+\frac{n}{2}(n-1) Q$ $\Rightarrow \frac{n}{2}[2 a+(n-1) d]=n P+\frac{n}{2}(n-1) Q$ $\Rightarrow 2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}=2 \mathrm{P}+(\mathrm{n}-1) \mathrm{Q}$ $\Rightarrow 2(\mathrm{a}-\mathrm{P})=(\mathrm{n}-1)(\mathrm{Q}-\mathrm...
Read More →For which of the following figures,
Question: For which of the following figures, diagonals are perpendicular to each other? (a) Parallelogram (b) Kite (c) Trapezium (d) Rectangle Solution: (b) The diagonals of a kite are perpendicular to each other....
Read More →In which of the following figures,
Question: In which of the following figures, all angles are equal? (a) Rectangle (b) Kite (c) Trapezium (d) Rhombus Solution: (a) In a rectangle, all angles are equal, i.e. all equal to 90....
Read More →For which of the following,
Question: For which of the following, diagonals bisect each other? (a) Square (b) Kite (c) Trapezium (d) Quadrilateral Solution: (a) We know that, the diagonals of a square bisect each other but the diagonals of kite, trapezium and quadrilateral do not bisect each other....
Read More →If three angles of a quadrilateral
Question: If three angles of a quadrilateral are each equal to 75o, the fourth angle is (a) 150o (b) 135o (c) 45o (d) 75o Solution: (b) 135o We know that, sum of interior angles of quadrilateral is equal to 360o. From the question it is given that, three angles of a quadrilateral are each equal to 75o. Let us assume the fourth angle be x. Then, 75o+ 75o+ 75o+ x = 360o 225 + x = 360o x = 360o 225 x = 135o...
Read More →Find the sum of all natural numbers from 1 and 100 which are divisible by 4 or 5.
Question: Find the sum of all natural numbers from 1 and 100 which are divisible by 4 or 5. Solution: To Find: The sum of all natural numbers from 1 to 100 which are divisible by 4 or 5. A number divisible by both 4 and 5 should be divisible by 20 which is the LCM of 4 and 5. Sum of numbers divisible by 4 OR 5 = Sum of numbers divisible by 4 + Sum of numbers divisible by 5 - Sum of numbers divisible by both 4 and 5. Sum of numbers divisible by $4=4+8+12+\ldots 100$ $=4(1+2+3+\ldots 25)=4 \times ...
Read More →If the sum of n terms of an AP is
Question: If the sum of n terms of an AP is $\left(3 n^{2}+5 n\right)$ and its mth term is 164, find the value of m. Solution: To Find: m Given: Sum of $n$ terms, $m^{\text {th }}$ term Put n = 1 to get the first term So $a_{1}=3+5=8$ Put n = 2 to get the sum of first and second term So $a_{1}+a_{2}=12+10=22$ So $a_{2}=14$ Common difference = 14 - 8 = 6 $T_{n}=a+(n-1) d=8+(n-1) 6=6 n+2$ Now 6m + 2 = 164 Or m = 27 The value of m is 27....
Read More →The sum of first 7 terms of an AP is 10 and that of next 7 terms is 17.
Question: The sum of first 7 terms of an AP is 10 and that of next 7 terms is 17. Find the AP. Solution: To Find: AP Given: Sum of first 7 terms $=10$ Sum of next 7 terms $=17$ According to the problem, Sum of first 14 terms of the given AP is $10+17=27$. So we can say $10=\frac{7}{2}(2 a+6 d)$ and $27=\frac{14}{2}(2 a+13 d)$ Solving the equations we get $14 a+42 d=20 \ldots$ (i) and $14 a+91 d=27 \ldots(i i)$ Subtracting (i) from (ii)we get 49d = 7 $\Rightarrow d=\frac{1}{7}$ Therefore from $(i...
Read More →Find the sum of all integers between 100 and 600, each of
Question: Find the sum of all integers between 100 and 600, each of which when divided by 5 leaves 2 as remainder. Solution: The integers between 100 and 600 divisible by 5 and leaves remainder 2 are 102, 107, 112, 117,, 597. To Find: Sum of the above AP Here a = 102, d = 5, l = 597 $a+(n-1) d=597$ $\Rightarrow 102+5(n-1)=597$ $\Rightarrow(n-1)=99$ $\Rightarrow n=100$ Now, $S=\frac{n}{2}[2 a+(n-1) d]$ $\Rightarrow S=\frac{100}{2}[2 \times 102+5(100-1)]$ $\Rightarrow S=50[204+495]=50 \times 699=3...
Read More →Find the sum of all integers between 101 and 500, which are divisible by 9.
Question: Find the sum of all integers between 101 and 500, which are divisible by 9. Solution: To Find: Sum of all integers between 101 and 500 divisible by 9 The integers between 101 and 500 divisible by 9 are $108,117,126, \ldots, 495$ (Add 9 to 108 to get 117,9 to 117 to get 126 and so on). Let a be the first term and d be the common difference and n be the number of terms of the AP Here a = 108, d = 9, l = 495 $\Rightarrow a+(n-1) d=495$ $\Rightarrow 108+9(n-1)=495$ $\Rightarrow 12+(n-1)=55...
Read More →Find the sum of all even integers between 101 and 199.
Question: Find the sum of all even integers between 101 and 199. Solution: To Find: The sum of all even integers between 101 and 199. The even integers form the following AP series - 102, 104, , 198 It is and AP series with a = 102 and l = 198. $198=102+(n-1) 2$ $\Rightarrow 96=(n-1) 2$ $\Rightarrow 48=n-1$ $\Rightarrow n=49$ Now, $S=\frac{49}{2}[102+198]=49 \times 150=7350$ The sum of all even integers between 101 and 199 is 7350....
Read More →Find the sum of all odd integers from 1 to 201.
Question: Find the sum of all odd integers from 1 to 201. Solution: To Find: The sum of all odd integers from 1 to 201. The odd integers form the following AP series: 1,3,5.201 First term = a = 1 Common difference = d = 2 Last term = 201 Let the number of terms be n $\Rightarrow 1+2(n-1)=201$ $\Rightarrow n-1=100$ $\Rightarrow n=101$ Sum of AP series $=\frac{n}{2}($ First term $+$ Last term $)$ $=\frac{101}{2}(1+201)$ = 101 101 = 10201 The sum of all odd integers from 1 to 201 is 10201....
Read More →If the ratio between the sums of n terms of two arithmetic progressions
Question: If the ratio between the sums of n terms of two arithmetic progressions is (7n + 1) : (4n + 27), find the ratio of their 11th terms. Solution: Given: Ratio of sum of $\mathrm{n}^{\text {th }}$ terms of 2 AP's To Find: Ratio of their $11^{\text {th }}$ terms Let us consider $2 \mathrm{AP}$ series $\mathrm{AP}_{1}$ and $\mathrm{AP}_{2}$. Putting n = 1, 2, 3 we get AP1 as 8, 15 22 and AP2 as 31, 35, 39. So, $a_{1}=8, d_{1}=7$ and $a_{2}=31, d_{2}=4$ For $\mathrm{AP}_{1}$ $S_{6}=8+(11-1) 7...
Read More →The sums of an terms of two arithmetic progressions are in the ratio
Question: The sums of an terms of two arithmetic progressions are in the ratio (7n 5) : (5n + 17). Show that their 6th terms are equal. Solution: Wrong question. It will be 7n + 5 instead of 7n 5. Given: Ratio of sum of n terms of 2 APs Let us consider $2 \mathrm{AP}$ series $\mathrm{AP}_{1}$ and $\mathrm{AP}_{2}$. Putting $n=1,2,3 \ldots$ we get $A P_{1}$ as $12,19,26 \ldots$ and $A P_{2}$ as $22,27,32 \ldots$ So, $a_{1}=12, d_{1}=7$ and $a_{2}=22, d_{2}=5$ For $\mathrm{AP}_{1}$ $S_{6}=12+(6-1)...
Read More →How many terms of the AP
Question: How many terms of the AP $20,19 \frac{1}{3}, 18 \frac{2}{3}$ must be taken to make the sum 300? Explain the double answer. Solution: To Find: Number of terms required to make the sum of the AP 300. Let the first term of the AP be a and the common difference be d Here $\mathrm{a}=20, d=-\frac{2}{3}$ $S_{n}=\frac{n}{2}[2 a+(n-1) d]$ $300=\frac{n}{2}\left[2 \times 20+(n-1)\left(-\frac{2}{3}\right)\right]$ $\Rightarrow 300 \times 6=n[120-2(n-1)]$ $\Rightarrow n[-2 n+122]=6 \times 300$ $\Ri...
Read More →How many terms of the AP 18 16, 14, 12
Question: How many terms of the AP 18 16, 14, 12, . are needed to give the sum 78? Explain the double answer. Solution: To Find: Number of terms required to make the sum 78. Here $a=18, d=-2$ Let n be the number of terms required to make the sum 78. $S_{n}=\frac{n}{2}[2 a+(n-1) d]$ $78=\frac{n}{2}[2 \times 18+(n-1)(-2)]$ $\Rightarrow 78 \times 2=36 n-2 n^{2}+2 n$ $\Rightarrow n^{2}-19 n+78=0$ $\Rightarrow n^{2}-6 n-13 n+78=0$ $\Rightarrow n(n-6)-13(n-6)=0$ $\Rightarrow(n-13)(n-6)=0$ Either n = 1...
Read More →How many terms of the AP 26, 21 16, 11,
Question: How many terms of the AP 26, 21 16, 11, . are needed to give the sum 11? Solution: To Find: Number of terms required Let the number of terms be n. $S_{n}=\frac{n}{2}[2 a+(n-1) d]$ $\Rightarrow \frac{n}{2}[52+(n-1)(-5)]=11 \Rightarrow \mathrm{n}[52-5 n+5]=22$ $\Rightarrow n(57-5 n)=11 \times 2=11[57-5(11)]$ $\Rightarrow \mathrm{n}=11$ 11 terms are required to give the sum 11 ....
Read More →If the sum of a certain number of terms of the
Question: If the sum of a certain number of terms of the AP 27, 24, 21, 18, . is 30, find the last term. Solution: To Find: Last term of the AP. Let the number of terms be n. $S_{n}=\frac{n}{2}[2 a+(n-1) d]$ $\Rightarrow \frac{n}{2}[54+(n-1)(-3)]=-30$ $\Rightarrow \mathrm{n}[54-3 \mathrm{n}+3]=-60$ $\Rightarrow 3 n^{2}-57 n-60=0$ $\Rightarrow n=\frac{57 \pm 63}{6}$ Either n = 20 or n = - 1 (n cannot be negative) Therefore n = 20 Also, $S=\frac{n}{2}(a+l)$ where l is the last term. $\Rightarrow-3...
Read More →Find the sum of n term of an AP whose
Question: Find the sum of $n$ term of an AP whose $r^{\text {th }}$ term is $(5 r+1)$ Solution: To Find: The sum of n terms of an AP Given: The $\mathrm{r}^{\text {th }}$ term. The $\mathrm{r}^{\text {th }}$ term of the series is given by $t_{r}=5 r+1$ Sum of the series is given by sum upto $n$ terms of $t_{r}$ $S_{r}=\sum_{i=1}^{n} t_{r}=\sum_{i=1}^{n} 5 r+1=\frac{5 n(n+1)}{2}+n$...
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