Question

If the sum of 18 terms of an

a.p. is 29 and the sum of its 29 terms is 18. then find the sum of its 47 terms.

Answer 1

We know that the general formula for the first n terms is give as:

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

Where symbols have their usual meanings.

Now, applying the formula to the given conditions we get:

Condition 1

$S_{18}=\frac{18}{2}[2a+17d]=29$

This can be simplified to $2a+17d=\frac{29}{9}$...........Equation 2

Likewise, let us begin with the next Condition.

Condition 2

$S_{29}=\frac{29}{2}[2a+28d]=18$

This can be simplified to as $2a+28d=\frac{36}{29}$...........Equation 3

Now, if we subtract Equation 1 from Equation 2, we will get:

$28d-17d=\frac{36}{29}-\frac{29}{9}$

$11d=-\frac{517}{261}$

$\therefore d=-\frac{517}{2871}$

Inserting this value of d in Equation 2, we get:

$2a=\frac{29}{9}-17\times \frac{-517}{2871} =\frac{1640}{261}$....Equation 4

Now, we know from the formula in equation 1 that the Sum of 47 terms will be:

$S_{47}=\frac{47}{2}[\frac{1640}{261}+46\times -\frac{517}{2871}]=-47$

Thus the sum of the 47 terms is -47