Find the sum of the first

In the given problem, we need to find the sum of terms for different arithmetic progressions. So, here we use the following formula for the sum of n terms of an A.P.,

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

(i) $2,6,10,14, \ldots$ To 11 terms.

Common difference of the A.P. $(d)=a_{2}-a_{1}$

= 6 − 2
= 4

Number of terms $(n)=11$

First term for the given A.P. $(a)=2$

So, using the formula we get,

$S_{n}=\frac{11}{2}[2(2)+(11-1)(4)]$

$=\left(\frac{11}{2}\right)[4+(10)(4)]$

$=\left(\frac{11}{2}\right)[4+40]$

$=\left(\frac{11}{2}\right)[44]$

$=242$

Therefore, the sum of first 11 terms for the given A.P. is.

(ii) $-6,0,6,12, \ldots$ To 13 terms.

Common difference of the A.P. $(d)=a_{2}-a_{1}$

= 0 − (−6)
= 6

Number of terms (n) = 13

First term for the given A.P. (a) = −6

So, using the formula we get,

$S_{n}=\frac{13}{2}[2(-6)+(13-1)(6)]$

$=\left(\frac{13}{2}\right)[-12+(12)(6)]$

$=\left(\frac{13}{2}\right)[-12+72]$

$=\left(\frac{13}{2}\right)[60]$

$=390$

Therefore, the sum of first 13 terms for the given A.P. is 390 .

(iii) 51 terms of an A.P whose $a_{2}=2$ and $a_{4}=8$

Now,

$a_{2}=a+d$

$2=a+d$ ....(1)

Also,

$a_{4}=a+3 d$

$8=a+3 d$.....(2)

Subtracting (1) from (2), we get

$2 d=6$

$d=3$

Further substituting in (1), we get

$2=a+3$

$a=-1$

Number of terms (n) = 51

First term for the given A.P. (a) = −1

So, using the formula we get,

$S_{\pi}=\frac{51}{2}[2(-1)+(51-1)(3)]$

$=\left(\frac{51}{2}\right)[-2+(50)(3)]$

$=\left(\frac{51}{2}\right)[-2+150]$

$=\left(\frac{51}{2}\right)[148]$

$=3774$

Therefore, the sum of first 51 terms for the given A.P. is 3774 .