# Find the sum of first 51 terms of an A.P. whose second

Question:

Find the sum of first 51 terms of an A.P. whose second and third terms are 14 and 18 respectively.

Solution:

In the given problem, let us take the first term as a and the common difference as d.

Here, we are given that,

$a_{2}=14$........(1)

$a_{3}=18$........(2)

Also, we know,

$a_{n}=a+(n-1) d$

For the $2^{\text {nd }}$ term $(n=2)$,

$a_{2}=a+(2-1) d$

$14=a+d$ (Using 1)

$a=14-d$....(3)

Similarly, for the 3rd term (n = 3),

$a_{3}=a+(3-1) d$

$18=a+2 d$ (Using 2)

$a=18-2 d$.......(4)

Subtracting (3) from (4), we get,

$a-a=(18-2 d)-(14-d)$

$0=18-2 d-14+d$

$0=4-d$

$d=4$

Now, to find a, we substitute the value of d in (4),

$a=14-4$

$a=10$

So, for the given A.P $d=4$ and $a=10$

So, to find the sum of first 51 terms of this A.P., we use the following formula for the sum of n terms of an A.P.,

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

Where; a = first term for the given A.P.

d = common difference of the given A.P.

= number of terms

So, using the formula for n = 51, we get,

$S_{51}=\frac{51}{2}[2(10)+(51-1)(4)]$

$=\frac{51}{2}[20+(50)(4)]$

$=\frac{51}{2}[20+200]$

$=\frac{51}{2}$

$=51(110)$

$=5610$

Therefore, the sum of first 51 terms for the given A.P. is $S_{51}=5610$.