Question

There are four terms in an A.P. In which sum of the first and last term is 13 and the

product of middle terms is 40. Then the terms of the given A.P​

Answer 1

Answer:

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Answer 2

The terms of the A.P are 2, 5, 8, and 11.

Given:

There are four terms in an A.P. The sum of the first and last term is 13.

The product of middle terms is 40.

To Find:

The terms of the given A.P​.

Solution:

If an A.P. contains four terms then they are given by $(a-3d)(a-d)(a+d)(a+3d)$.

As the sum of the first and last term is 13. So,

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

It is given that the product of middle terms is 40. So,

$(a-d)(a+d)=40$

Remove the brackets by multiplying the terms.

$a^2-ad+ad-d^2=40\\a^2-d^2=40$

Substitute $a=\frac{13}{2}$ in $a^2-d^2=40$ and find d.

$(\frac{13}{2}) ^2-d^2=40\\\frac{169}{4}-d^2=40\\ d^2=\frac{169}{4}-40\\=\frac{169-160}{4}\\ =\frac{9}{4}\\d=\frac{3}{2}$

Substitute $a=\frac{13}{2}$ and $d=\frac{3}{2}$ in $(a-3d)(a-d)(a+d)(a+3d)$and find the terms of A.P.

$(a-3d)=\frac{13}{2}-\frac{3\cdot 3}{2}\\=\frac{13-9}{2}\\ =\frac{4}{2}\\ =2$

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

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

$(a+3d)=\frac{13}{2}+\frac{3\cdot 3}{2}\\=\frac{13+9}{2}\\ =\frac{22}{2}\\ =11$

Thus, the terms of the A.P are 2, 5, 8, and 11.