Name: Finding the nth Term of A.P. With Given Two Terms
Uploaded: 2019-07-19
Duration: 2 min 38 s
Description: Add the common difference to the first term to get the second term. Add the common difference to the second term to get the third term. Continue to identify...

Question

Find the sum of first

51

terms of an AP whose second and third terms are

14

and

18

respectively.

Accepted Answer

In the given problem, let us take the first term as $a$ and the common difference as $d$ .
Here, we are given that, $a_{2} = 14$ and $a_{3} = 18$ .

Also, we know, $a_{n} = a + (n - 1) d$ .

For second term, $a_{2} = a + (2 - 1) d$

$14 = a + d$

$a = 14 - d \dots \dots (1)$

Similarly, for the third term, $a_{3} = a + (3 - 1) d$ .

$a = 18 - 2 d \dots \dots (2)$

Subtracting $(1)$ from $(2)$ , 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 $(2)$ ,

$a = 14 - 4$

$a = 10$

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 form for the sum of $n$ terms of an A.P.,

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

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

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

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

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

$= \frac{51}{2} [220]$

$= 51 (110)$

$= 5610$

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

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

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