{7^{th}} term of an A.P. is 40 , then sum of first 13 terms is

English

Gujarati

Hindi

8. Sequences and Series

easy

${7^{th}}$ term of an $A.P.$ is $40$, then the sum of first $13$ terms is

A

$53$

B

$520$

C

$1040$

D

$2080$

Solution

(b) ${7^{th}}$ term of an $A.P. $$ = 40$

$a + 6d = 40$

${S_{13}} = \frac{{13}}{2}[2a + (13 – 1)d] = \frac{{13}}{2}[2(a + 6d)]$

$ = \frac{{13}}{2}\;.\;2\;.\;40 = 520$.

Standard 11

Mathematics

Share

Similar Questions

The difference between any two consecutive interior angles of a polygon is $5^{\circ}$ If the smallest angle is $120^{\circ},$ find the number of the sides of the polygon.

hard

A farmer buys a used tractor for $Rs$ $12000 .$ He pays $Rs$ $6000$ cash and agrees to pay the balance in annual instalments of $Rs$ $500$ plus $12 \%$ interest on the unpaid amount. How much will the tractor cost him?

hard

Consider an $A.P.$ of positive integers, whose sum of the first three terms is $54$ and the sum of the first twenty terms lies between $1600$ and $1800$ . Then its $11^{\text {th }}$ term is :

hard

(JEE MAIN-2025)

If $19^{th}$ terms of non -zero $A.P.$ is zero, then its ($49^{th}$ term) : ($29^{th}$ term) is

hard

(JEE MAIN-2019)

If the ${p^{th}}$ term of an $A.P.$ be $q$ and ${q^{th}}$ term be $p$, then its ${r^{th}}$ term will be

easy