Question

Find three terms of an AP whose sum and products will be 21 and 231

Answer 1

Answer:

3, 7, 11

Step-by-step explanation:

Suppose 3 terms of an AP are,

a-d, a , a+d

Sum of these terms is 21,

so a-d+a+a+d = 21

3a = 21 (after cutting +ve and -ve d)

a = 7

Now, product of these terms is 231

so (a-d)(a)(a+d) = 231

a($a^{2} - d^{2}$) = 231 (using identity (a-b)(a+b)=($a^{2} - b^{2}$))

now put the value of a = 7 we have,

7 (49 - ($d^{2}$)) = 231

(49 - ($d^{2}$)) = 231/7

(49 - ($d^{2}$)) = 33

49 - 33 = ($d^{2}$)

16 = ($d^{2}$)

so d = + 4 or -4 ($\sqrt{16}$ = ± 4 )

For a = 7 and d = 4

fisrt term = a - d = 7 - 4 = 3

second term = a = 7

third term = a + d = 7 + 4 = 11

For a = 7 and d = - 4

fisrt term = a - d = 7 - (-4) = 7 + 4 = 11

second term = a = 7

third term = a + d = 7 + (-4) = 7 - 4 = 3

So our thee terms of an AP are 3, 7, 11