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.

Answer 1

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.,

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

d = common difference of the given A.P.

n = number of terms

(i) To 11 terms.

Common difference of the A.P. (d) =

= 6 − 2
= 4

Number of terms (n) = 11

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

So, using the formula we get,

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

(ii) To 13 terms.

Common difference of the A.P. (d) =

= 0 − (−6)
= 6

Number of terms (n) = 13

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

So, using the formula we get,

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

(iii) 51 terms of an A.P whose and

Now,

Also,

$$\begin{gathered} a_4=a+3d \\ 8=a+3d...\left(2\right) \end{gathered}$$

Subtracting (1) from (2), we get

Further substituting in (1), we get

Number of terms (n) = 51

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

So, using the formula we get,

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