Question

ther are four terms in an ap in which sum of the first and last term is 13 and the product of middle term is 40 then the terms of the given ap​

Answer 1

Answer:

Please refer to attached picture...

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Answer 2

Step-by-step explanation:

Given:The first and last term is 13 and the product of middle terms is 40.

To find: The terms of the given A.P.

Solution:

Tip: It is always convenient to take four terms of A.P. as a-3d, a-d,a+d and a+3d.

Let the four terms of AP are

a-3d, a-d,a+d and a+3d

ATQ

Sum of the first and last term is 13;

$a - 3d + a + 3d = 13 \\ \\ 2a = 13 \\ \\ \bold{\green{a = \frac{13}{2} }}$

Product of middle terms is 40;

$(a - d)(a + d) = 40 \\ \\ {a}^{2} - {d}^{2} = 40$

put the value of a

$\frac{169}{4} - {d}^{2} = 40 \\ \\ {d}^{2} = \frac{169}{4} - 40 \\ \\ {d}^{2} = \frac{169 - 160}{4} \\ \\ {d}^{2} = \frac{9}{4} \\ \\ or \\ \\ \bold{\red{d = ± \frac{3}{2} }}$

Thus, terms of A.P. are,taking positive value of d;

$a - 3d = \frac{13}{2} - 3 \times \frac{3}{2} \\ \\ a - 3d = \frac{13}{2} - \frac{9}{2} \\ \\ a - 3d = \frac{4}{2} \\ \\ a - 3d = 2$

and

$a - d = \frac{13}{2} - \frac{3}{2} \\ \\ a - d = \frac{10}{2} \\ \\ a - d = 5$

and

$a + d = \frac{13}{2} + \frac{3}{2} \\ \\ a + d = \frac{13 + 3}{2} \\ \\ a + d = \frac{16}{2} \\ \\ a + d = 8$

and

$a + 3d = \frac{13}{2} + 3 \times \frac{3}{2} \\ \\ a + 3d = \frac{13}{2} + \frac{9}{2} \\ \\ a + 3d = \frac{22}{2} \\ \\ a + 3d = 11$

Final answer:

Terms of A.P. are 2,5,8 and 11.

or can be 11,8,5 and 2 if negative value of d will be consider.

Verification:

Sum of the first and last term is 2+11= 13;

Product of middle two 8×5=40.

Hope it helps you.

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