Question

The sum of four consecutive terms which are in an arithmetic progression 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 number.

Answer 1

Hint: In an arithmetic progression the consecutive terms differ by a common difference. From convention, the first term of arithmetic progression (A.P) is a and the common difference is d, which can be found by subtracting any two consecutive terms.

Complete step-by-step answer:
Let the four consecutive terms of an A.P be a – 3d, a – d, a + d, a + 3d, with common difference 2d.
Now, according to the question,
a – 3d + a – d + a + d + a + 3d = 32
Therefore, 4a = 32.
a = 8
so, the terms become 8 – 3d, 8 – d, 8 + d, 8 + 3d. Now, again according to the question,
$\dfrac{{(8 - 3d)(8 + 3d)}}{{(8 - d)(8 + d)}} = \dfrac{7}{{15}}$
Now, we will use the property \[(a - b)(a + b) = {a^2} - {b^2}\] to simplify numerator and denominator, so we get
$\dfrac{{({8^2} - {{(3d)}^2})}}{{({8^2} - {d^2})}} = \dfrac{7}{{15}}$
Now, we will do cross – multiplication.
$15(64 - 9{d^2}) = 7(64 - {d^2})$
Solving this above equation, we get
$15(64) - 15(9{d^2}) = 7(64) - 7{d^2}$
Moving like terms on same side, we get
$15(64) - 7(64) = 15(9{d^2}) - 7{d^2}$
$8(64) = 128{d^2}$
${d^2} = 4$
$d = \pm 2$
Now, when d = 2,
So, the terms are 2, 6, 10, 14
When d = -2,
The terms are 14, 10, 6, 2.

Note: Whenever we come up with such types of questions, we will let the terms such that the calculations become easy. Like in this question, if we take a, a + d, a + 2d, a + 3d as the terms, then the calculations are very lengthy and you may also make some errors while calculating the values of a and d. The terms which you let should have an equal common difference. Also, use property \[(a - b)(a + b) = {a^2} - {b^2}\] as required. Using such properties helps us in solving these questions without committing a mistake.