Find four consecutive terms of an ap such that the difference if the middle terms is 10 and the product of the extreme terms is 216

Open in App

Solution

dear student

Let the terms be

a,a+d,a+2d,a+3d
according to the question
(a+2d)-(a+d)=10

a+2d-a-d=10

d=10
now

(a)(a+3d)=216
$a^{2} + 3 a d = 216 a^{2} + 30 a = 216 a^{2} + 30 a - 216 = 0 a^{2} + 36 a - 6 a - 261 = 0 a (a + 36) - 6 (a + 36) = 0 (a - 6) (a + 36) = 0 s o a = 6, - 36$
if a=6 then ap= 6,16,26,36
if a=-36 the ap=-36,-26,-16,-6

regards

Suggest Corrections

Similar questions

Q. The sum of four consecutive terms of an

A P

32

and the ratio of the product of the
first and the last terms to the product of the two middle terms is

7 : 15.

Find the numbers.

Q. Sum of three consecutive terms of an AP is 24 and product is 312. Find the middle term and common diffeerence of the AP.

The sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and the last term to the product of two middle terms is 7/ 15 . Find the numbers

Sum of four consecutive number in an AP is 32 and the ratio of products of first and last term to the product of middle term is 7:15. Find the no.

The sum of four consecutive numbers in an AP is 32. The ratio of the product of the first and the last term to the product of the two middle terms is 7:15 . Find the numbers

Join BYJU'S Learning Program

Find four consecutive terms of an ap such that the difference if the middle terms is 10 and the product of the extreme terms is 216

dear student Let the terms be a,a+d,a+2d,a+3d according to the question (a+2d)-(a+d)=10 a+2d-a-d=10 d=10 now (a)(a+3d)=216 a2+3ad=216a2+30a=216a2+30a-216=0a2+36a-6a-261=0a(a+36)-6(a+36)=0(a-6)(a+36)=0so a=6,-36 if a=6 then ap= 6,16,26,36 if a=-36 the ap=-36,-26,-16,-6 regards