# If the sum of three consecutive terms of an increasing A.P.

Question:

If the sum of three consecutive terms of an increasing A.P. is 51 and the product of the first and third of these terms is 273, then the third term is

(a) 13

(b) 9

(c) 21

(d) 17

Solution:

In the given problem, the sum of three consecutive terms of an A.P is 51 and the product of the first and the third terms is 273.

We need to find the third term.

Here,

Let the three terms be where, a is the first term and d is the common difference of the A.P

So,

$(a-d)+a+(a+d)=51$

$3 a=51$

$a=\frac{51}{3}$

$a=17$

Also,

$(a-d)(a+d)=273$

$a^{2}-d^{2}=273 \quad\left[U \sin g a^{2}-b^{2}=(a+b)(a-b)\right]$

$17^{2}-d^{2}=273$

$289-d^{2}=273$

Further solving for d,

Now, it is given that this is an increasing A.P. so cannot be negative.

So, d = 4

Substituting the values of a and d in the expression for the third term, we get,

Third term =

So,

$a+d=17+4$

$=21$

Therefore, the third term is $a_{3}=21$

Hence, the correct option is (c).