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Sum of the first 14 terms of an A.P is 1505 and its first term is 10. Find its 25th term.

Answer
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Hint: To find the 25th term we need the first term and the common difference from the given information. We can find the nth term of an A.P. by using the following formula ${{a}_{n}}=a+(n-1)d$ .

Complete step-by-step answer:
To find the 25th term, we have n=25 and we need to find a and d i.e. first term and common difference of the A.P.
The given information is the sum of the first 14 terms of an A.P.
The sum of first n terms of an A.P. is given by ${{S}_{n}}=\dfrac{n}{2}\{2a+(n-1)d\}$ ……… (i) Where again a=first term and d=common difference of the A.P.
We substitute the value of n=14 and a=10 in equation (i)
${{S}_{14}}=\dfrac{14}{2}\{2.10+(14-1)d\}$

From the question we have, ${{S}_{14}}=1505$ , therefore by substituting the value of ${{S}_{14}}$ in the above equation we have,
$1505=\dfrac{14}{2}\{2.10+(14-1)d\}$ ……………. (ii)
From the above equation we can find the value of d
Now we simplify equation (ii)
$1505=7(20+13d)$
Dividing LHS and RHS by 7 we have,
$\begin{align}
 & \dfrac{1505}{7}=20+13d \\
 & \Rightarrow 215=20+13d \\
\end{align}$

Subtracting 20 both sides we have,
$\begin{align}
 & \Rightarrow 195=13d \\
 & \Rightarrow d=\dfrac{195}{13} \\
 & \Rightarrow d=15 \\
\end{align}$

Therefore, now we have the value of d is equal to 15.
Now we can find the 25th term as we have n, a and d for the formula ${{a}_{n}}=a+(n-1)d$ .
a=10, d=15 and n=25
The 25th term is given by,
$\begin{align}
 & {{a}_{25}}=a+(n-1)d \\
 & =10+(25-1)15 \\
\end{align}$

On further simplification we have,
$\begin{align}
 & {{a}_{25}}=10+24\times 15 \\
 & =10+360 \\
 & =370 \\
\end{align}$
Hence, the answer is 370.

Note: For solving questions of an A.P. there are two variables first term, common difference and nth term. We need all of these to solve any problem of an A.P and for this we need three equations for these three variables. For this question two variables i.e. first term and n were already given to us and one equation was also given to find common differences.