Question

three terms are there in Ap whose sum is 21 and their product is 231 find the term




plzz solve it's imp for my exam​

Answer 1

[Always remember this]

To find 3 terms in an AP

take these terms---- (a-d), a, (a+d)

to find 4 terms----- (a-3d), (a-d), (a+d), (a+3d)

to find 5 terms----- (a-2d), (a-d), a, (a+d), (a+2d)

to find 6 terms----- (a-5d), (a-3d), (a-d), (a+d), (a+3d), (a+5d)

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NOW, LET'S SOLVE YOUR QUESTION

let the three terms are :-

(a-d), a, (a+d)

case 1 :-

sum of these terms = 21

(a - d) + (a) + (a+d) = 21

a - d + a + a + d = 21

d cancels out

3a = 21

a = 7

Case 2 :-

product of these terms = 231

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

$a(a^{2} - d^{2}) = 231$

put the value of a

$7(7^{2} - d^{2})= 231\\49 - d^{2} = 231/7\\49 - d^{2} = 33\\d^{2} = 49 - 33\\d^{2} = 16\\d = \sqrt{16}\\d= 4$

Now, we got both 'a' and 'd'

(a-d) = 7-4 = 3

a = 7

(a+d) = 7+4 = 11

Hence, the terms are :-

3, 7, 11