Question

The number of terms in an

A . P

is even, the sum of odd terms is

24

and the sum of even terms is

30

. If the last term exceeds the first term by

\frac{21}{2}

,find the number of terms in the

A . P

.

Accepted Answer

Let $a$ be the first term and $d$ be the common difference of $A . P$ .

Let the number of terms of the given $A . P$ be $2 n$ .

Hence there are $n$ even terms and $n$ odd terms according to given condition, we get

$a_{1} + a_{3} + a_{5} \dots \dots n terms = 24$

$\Rightarrow a + a + 2 d + a + 4 d \dots . . . n terms = 24$

We know that, sum of $n$ terms of an $A . P .$ is $S_{n} = [2 a + (n - 1) d]$ where $a$ is the first term and $d$ is the common difference.

$\Rightarrow \frac{n}{2} [2 a + (n - 1) 2 d] = 24$

$\Rightarrow n [a + (n - 1) d] = 24$ (i)

Also, $a_{2} + a_{4} + a_{6} \dots n terms = 30$

We know that, $n^{t h}$ term of an $A . P .$ is $a_{n} = a + (n - 1) d$

$\Rightarrow (a + d) + (a + 3 d) + (a + 5 d) \dots n terms = 30$

We know that, sum of $n$ terms of an $A . P .$ is $S_{n} = [2 a + (n - 1) d]$ where $a$ is the first term and $d$ is the common difference.

$∴ \frac{n}{2} [2 (a + d) + (n - 1) 2 d] = 30$

$\Rightarrow n [a + d + (n - 1) d] = 30$

$\Rightarrow n [a + n d] = 30$ (ii)

Also, $a_{2 n} = a_{1} + \frac{21}{2}$

$\Rightarrow a + (2 n - 1) d = a + \frac{21}{2}$

$\Rightarrow (2 n - 1) d = \frac{21}{2}$ (iii)

Consider, equation (i) $-$ equation (ii)

$[n (n - 1) - n^{2}] d = 24 - 30$

$\Rightarrow n d = 6$

On dividing equation (iii) by equation (iv); we get

$\frac{2 n - 1}{n} = \frac{21}{12}$

$\Rightarrow \frac{2 n - 1}{n} = \frac{7}{4}$

$\Rightarrow 8 n - 4 - 7 n \Rightarrow n = 4$

Hence, the required number of terms of given $A . P$ be $2 n$ $i.e. 2 \times 4 i.e. 8$ .

M L Aggarwal Solutions for Chapter: Sequences and Series, Exercise 10: EXERCISE 10.10

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Sequences and Series, Exercise 10: EXERCISE 10.10

Questions from M L Aggarwal Solutions for Chapter: Sequences and Series, Exercise 10: EXERCISE 10.10 with Hints & Solutions

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