Find the sum of the first
Question:

Find the sum of the first

(i) 11 terms of the A.P.2, 6, 10. 14

(ii) 13 terms of the A.P. −6, 0, 6, 12, …

(iii) 51 terms of the A.P. : whose second term is 2 and fourth term is 8.

Solution:

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.

= 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 .

 

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