Find the sum of first 22 terms of an A.P. in which d = 22 and a22 = 149.

Question:

Find the sum of first 22 terms of an A.P. in which d = 22 and a22 = 149.

Solution:

In the given problem, we need to find the sum of first 22 terms of an A.P. Let us take the first term as a.

Here, we are given that,

$a_{22}=149$...........(1)

$d=22$...............(2)

Also, we know,

$a_{n}=a+(n-1) d$

For the 22nd term (n = 22),'

$a_{22}=a+(22-1) d$

$149=a+21(22) \quad$ (Using 1 and 2 )

$a=149-462$

 

$a=-313$..........(3)

So, as we know the formula for the sum of n terms of an A.P. is given by,

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

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

d = common difference of the given A.P.

= number of terms

So, using the formula for n = 22, we get,

$S_{22}=\frac{22}{2}[2(-313)+(22-1)(22)]$

$=(11)[-626+(21)(22)] \quad$ (Using 2 and 3 )

$=(11)[-626+462]$

$=(11)[-164]$

 

$=-1804$

Therefore, the sum of first 22 terms for the given A.P. is $S_{22}=-1804$.

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