Question

Find the value of the middle term of the following AP: -6,-2, 2 .............58.

OR

Determine the AP whose fourth term is 18 and the differences of the ninth term from the fifteenth term is 30.

Answer 1

The given AP is -6, -2, 2, ....., 58

Here, first term a = - 6 and common difference d = -2 -(-6) = -2 + 6 = 4

Last term, l = 58

$\Rightarrow$ a+ (n -1)d = 58

$\Rightarrow$ -6 + (n -1)$\times$ 4 = 58

$\Rightarrow$ (n-1) × 4 = 64

$\Rightarrow$ (n-1) =16

$\Rightarrow$ n = 17

Middle term of the A.P. ${\left(\frac{n+1}{2}\right)}^{th}term={\left(\frac{17+1}{2}\right)}^{th}term=9^{th}term$

$a_9$ = a+ (9-1) d = -6 + 8 $\times$ 4 = -6 +32 = 26

Thus, the middle term of the given A.P. is 26.

OR

Let the first term of the given AP be 'a' and the common difference be 'd'.

We have $a_4$ = 18

a+(4-1)d = 18

$\Rightarrow$ a+3d = 18 ..............(i)

Also, it is given that

$a_{15}-a_9=30$
$\Rightarrow$ a + (15 - 1) d - {a + (9 - 1) d} = 30

$\Rightarrow$ a +14d - (a + 8d) = 30

$\Rightarrow$ 6d = 30

$\Rightarrow$ d = 5

Putting the value of d in (i):

a+ 3 × 5 = 18

$\Rightarrow$ a + 15 = 18

$\Rightarrow$ a = 18 -15

$\Rightarrow$ a =3

Therefore, the first term and the common difference of the AP are 3 and 5 respectively

Thus, the A.P. is 3, 3+5, 3+ (2$\times$5), 3+ (3$\times$5) .........

That is 3, 8, 13, 18...