Question

IF 3rd and 7th term of an A.P
is
18 and 30 then find the sum
of first 17 terms of and A.P.
b) 612
c) 624
d) 636​

Answer 1

We know that :-

$\boxed{ \boxed{\tt \large T_n = a+(n-1)d} } \\ \\ \tt where : \\ \\ \tt \bull T_n = n_{th} \: term \\ \\ \tt \bull n = number \: of \: terms \\ \\ \tt \bull a= first \: term \\ \\ \tt \bull d = common \: difference$

$\tt \rightarrow 3rd \: term : \\ \\ \tt\longrightarrow a + (3-1)d = 18 \\ \\ \tt \longrightarrow a + 2d = 18\\ \\ \tt \longrightarrow a = 18 - 2d.....(1)$

$\tt \rightarrow 7th \: term : \\ \\ \tt \longrightarrow a + ( 7-1)d = 30 \\ \\ \tt \longrightarrow a + 6d= 30\\ \\ \tt \longrightarrow a = 30 - 6d..........(2)$

Now, from equation (1) and (2) :-

$\tt \longrightarrow 18 - 2d = 30 - 6d$

$\tt \longrightarrow 6d - 2d = 30 - 18$

$\tt \longrightarrow 4d = 12$

$\tt \longrightarrow d = 3$

Now , to find out the value of a put d = 3 in equation (1).

$\tt \longrightarrow a = 18 - 2(3) \\ \\ \tt \longrightarrow a = 18 - 6 \\ \\ \tt \longrightarrow a = 12$

Therefore , Value of a = 12 and value of d = 3

Now , we have to find out sum of first 17 terms.

We know that :-

$\boxed{ \boxed{\tt \large s_n = \frac{n}{2} (2a+(n-1)d)} } \\ \\ \tt where : \\ \\ \tt \bull s_n =sum \: of \: n\: term \\ \\ \tt \bull n = number \: of \: terms \\ \\ \tt \bull a= first \: term \\ \\ \tt \bull d = common \: difference$

$\tt \longrightarrow s_{17} = \dfrac{17}{2} (24+48) \\ \\ \tt \longrightarrow s_{17} = \dfrac{17}{2} \times 72 \\ \\ \tt\longrightarrow s_{17} = 17 \times 36\\ \\ \tt\longrightarrow s_{17} = 612$

Answer :

Option b)612 is correct