Question

Find 6 consecutive terms in an A.P. whose sum is 72 and the product of first and last term is 44

Answer 1

Answer:

2, 6, 10, 14, 18, 22

Step-by-step explanation:

Hi,

Let the 6 consecutive terms of an AP be

a-5d, a-3d, a-d, a+d, a+3d, a+5d where '2d' was common difference.

Given, sum = 72

=> 6a = 72

=> a = 12.

Also given , product of first and last term is 44

=> a² - 25d² = 44

=>12² - 25d² =44

=> 25d² = 100

=>d² = 4

=> d = ±2.

Hence, the 6 term of AP are

              2, 6, 10, 14, 18, 22.

Hope, it helped !

Answer 2

➡️Answer: 2,6,10,14,18,22

➡️Solution:

To find 6 consecutive terms in an A.P. whose sum is 72,

let the terms are

➡️a,a+d,a+2d,a+3d, a+4d, a+5d

Now their sum

$a + a + d + a + 2d +a + 3d +a +4 d +a +5 d = 72 \\ \\ 6a + 15d = 72 \\ \\ 2a + 5d = 24 \: \: \: eq1 \\ \\ 5d = 24 - 2a \: \: \: eq2$
Product of first and last

$a(a + 5d) = 44 \\ \\ {a}^{2} + 5da = 44$
put the value of 5d from eq2
${a}^{2} + a({24 - 2a}) = 44 \\ \\ {a}^{2} + 24a - 2{a}^{2} = 44 \\ \\ - {a}^{2} + 24a - 44 = 0 \\ \\ {a}^{2} - 24a + 44 = 0 \\ \\ {a}^{2} - 22a - 2a + 44 = 0 \\ \\ a(a - 22) - 2(a - 22) = 0 \\ \\ (a - 22)(a - 2) = 0 \\ \\ a = 22 \\ \\ a = 2\\ \\so \\ \\ 2a + 5d = 24 \\ \\ 2(2) + 5d = 24 \\ \\ 5d = 20 \\ \\ d = 4 \\ \\ or \: a = 22 \\ \\ 5d = 24 - 44 \\ \\ 5d = - 20 \\ \\ d = - 4$
So AP is when (a= 2,d = 4)

2,6,10,14,18,22

or

(a=22,d=-4)

22,18,14,10,6,2

By inspecting both are same

Hope it helps you.