If the 7th term of A.P is 49 and the 17th term of A.P is 289, find the

1.	If the 7th term of A.P is 49 and the 17th term of A.P is 289, find the sum of n term.
Answer» Answer: let 7th TERM 49 $= a_{7}=a+(7-1)d$ ....EQN(1)⇒ $49=a+6d$ the 17th term 289 $= a_{17} =a+(17-1)d$ ....eqn(2)⇒ $289=a+16d$ solving eqn 1 and 2 simultaneously⇒ subtract (2) from(1)⇒ $289-49=16d-6d$ ⇒ $240=10d$ ⇒ ∴ $d=24$ putting d=24 in EQUATION 1 ⇒ $49=a+6(24)$ ⇒ $49=a+144$ ⇒ $49-144=a$ ⇒ ∴ $-95=a$ the nth term $S_{n} =\frac{n}{2} \left[\begin{array}{ccc}2a+(n-1)d\end{array}\right] =$ $S_{n} =\frac{n}{2} \left[\begin{array}{ccc}2(-95)+(n-1)24\end{array}\right] =$ ⇒ $S_{n} =\frac{n}{2} \left[\begin{array}{ccc}-190+24n-24\end{array}\right]$ ⇒ $S_{n} =\frac{n}{2} \left[\begin{array}{ccc}-190-24+24n\end{array}\right]$ ⇒ $S_{n} =\frac{n}{2} \left[\begin{array}{ccc}-214+24n\end{array}\right]$ ⇒ $S_{n} =n(-107+12n)$ ⇒ $S_{n} =-107n+12n^{2}$ as the SUM of the n term Step-by-step explanation:

If the 7th term of A.P is 49 and the 17th term of A.P is 289, find the sum of n term.

Answer»

Answer:

let 7th TERM 49 $= a_{7}=a+(7-1)d$ ....EQN(1)⇒ $49=a+6d$

the 17th term 289 $= a_{17} =a+(17-1)d$ ....eqn(2)⇒ $289=a+16d$

solving eqn 1 and 2 simultaneously⇒

subtract (2) from(1)⇒ $289-49=16d-6d$ ⇒

$240=10d$ ⇒

∴ $d=24$

putting d=24 in EQUATION 1 ⇒

$49=a+6(24)$ ⇒

$49=a+144$ ⇒

$49-144=a$ ⇒

∴ $-95=a$

the nth term $S_{n} =\frac{n}{2} \left[\begin{array}{ccc}2a+(n-1)d\end{array}\right] =$ $S_{n} =\frac{n}{2} \left[\begin{array}{ccc}2(-95)+(n-1)24\end{array}\right] =$ ⇒