#### Question

Find the A.P. whose fourth term is 9 and the sum of its sixth term and thirteenth term is 40.

#### Solution

Let the first term be *a* and the common difference be *d*.

It is given that *a*_{4} = 9 and *a*_{6} + *a*_{13} = 40.

*a*_{4} = 9

`rArr a+(4-1)d=9` `a_n=a+(n-1)d`

`rArr a+3d=9`

`a_6+a_13=9`

`rArr {a+(6-1)d}+{a+(13-1)d}=40`

`rArr {a+5d}+{a+12d}=40`

`rArr 2a+17d=40`

From (1):

*a* = 9 − 3*d*

Substituting the value of *a* in (2):

2 (9 − 3*d*) + 17*d* = 40

⇒ 18 + 11*d* = 40

⇒ 11*d* = 22

⇒ *d* = 2

∴ *a* = 9 − 3 × 2 = 3

Thus, the given A.P. is *a*, *a* + *d*, *a* + 2*d *…, where *a* = 3 and *d* = 2.

Thus, the A.P. is 3, 5, 7, 9 …

Is there an error in this question or solution?

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Find the A.P. Whose Fourth Term is 9 and the Sum of Its Sixth Term and Thirteenth Term is 40. Concept: Sum of First n Terms of an AP.

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