Question

T5:t3=5:2 then find the value of t15:t7

Answer 1

The value of  $t_15:t_7$  = 20 : 8 or 5 : 2

Step-by-step explanation:

We have,

$t_5:t_3$  = 5 : 2

To find, the value of $t_15:t_7$=

Let a, d be the first term and common difference respectively.

The nth term of an AP,

$t_n=a+(n-1)d$

∴$t_5:t_3=5:2$

⇒ $\dfrac{t_5}{t_3}=\dfrac{5}{2}$

⇒ $\dfrac{a+4d}{a+2d}=\dfrac{5}{2}$

$a+4d=5$              ..... (1)

and a + 2d = 2      ......(2)

Subtracting (1) from (2), we get

a + 4d - a - 2d = 5 - 2

⇒ 2d = 3

⇒ d=$\dfrac{3}{2}$

Put d = $\dfrac{3}{2}$ in (1), we get

$a +4(\dfrac{3}{2})= 5$

⇒ a = 5 - 6 = - 1

∴ $t_15$ = a + 14d

=$-1+14(\dfrac{3}{2})$= - 1 + 21 = 20

$t_15$ = 20

and $t_7$ = a + 6d

=-1+6$(\dfrac{3}{2})$=-1+9

$t_7$= 8

∴ The value of  = 20 : 8 or 5 : 2