Question

Find the sum of first 30 terms of an AP whose third and fourth terms are 19 and 22 respectively.​

Answer 1

Answer:1695

Step-by-step explanation:

a+2d=19-(1)

a+3d=22-(2)

(2)-(1)

d=3

a+2d=19

a+6=19

a=13

x30=a+29d

=13+(29×3)

=13+87

=100

sum of thirty term

s30=n/2(a+x30)

=30/2(13+100)

=15×113

=1695

Answer 2

Answer:-

$\sf{The \ sum \ of \ first \ 30 \ terms \ is \ 1695.}$

Given:

• $\sf{t_{3}=19}$
• $\sf{t_{4}=22}$

To find:

$\sf{Sum \ of \ first \ 30 \ terms.}$

Solution:

$\boxed{\sf{tn=a+(n-1)d}}$

$\sf{According \ to \ first \ condition. }$

$\sf{a+2d=19...(1)}$

$\sf{According \ to \ second \ condition. }$

$\sf{a+3d=22...(2)}$

$\sf{But, \ t_{4}=t_{3}+d}$

$\sf{\therefore{22=19+d}}$

$\sf{\therefore{d=22-19}}$

$\boxed{\sf{\therefore{d=3}}}$

$\sf{Substitute \ d=3 \ in \ equation(1), \ we \ get}$

$\sf{a+2(3)=19}$

$\sf{\therefore{a+6=19}}$

$\sf{\therefore{a=19-6}}$

$\boxed{\sf{\therefore{a=13}}}$

_______________________________________

$\sf{Here, \ a=13 \ and \ d=3}$

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

$\sf{\therefore{S_{30}=\frac{30}{2}[2(13)+(30-1)\times3]}}$

$\sf{\therefore{S_{30}=15[26+29\times3]}}$

$\sf{\therefore{S_{30}=15[26+87]}}$

$\sf{\therefore{S_{30}=15\times113}}$

$\sf{\therefore{S_{30}=1695}}$

$\sf\purple{\tt{\therefore{The \ sum \ of \ first \ 30 \ terms \ is \ 1695.}}}$