Question

The sum of first seven terms of an A.P. is 182. If its 4th and the 17th terms are in the ratio 1 : 5, find the A.P.

Answer 1

GIVEN :

Sum of first seven terms of an AP = 182

4th term of an AP and 17th term are in ratio 1:5

4th term = a + 3d

17th term = a + 16d

$\frac{a + 3d}{a + 16d} = \frac{1}{5} \\ \\ a + 16d = 5(a + 3d) \\ \\ a + 16d = 5a + 15d \\ \\ 16d - 15d = 5a - a \\ \\ d = 4a - - - (1)$

In an AP sum of the terms = n/2 ( 2a + (n - 1)d )

182 = 7/2 ( 2a + (7 - 1)d)

182 = 7/2 ( 2a + 6d )

182 = 7/2 ( 2a + 6d )

182 × 2 = 7( 2a + 6d )

364 = 7 ( 2a + 6d )

364/7 = 2a + 6d

52 = 2a + 6 ( 4a)

52 = 2a + 24a

52 = 26a

a = 52/26

a = 2

First term = 2

Substitute a in eq - (1)

d = 4(2)

d = 8

Common Difference = 8

Second term = a + d = 2 + 8 = 10

Third Term = a + 2d = 2 + 16 = 18

Fourth term = a + 3d = 2 + 24 = 26

Therefore, the AP is 2, 10, 18 , 26.......

Answer 2

• Sum of first seven terms of an A.P. is 182

$S_{7}$ = 182 _____ [GIVEN]

$S_{n}$ = $\dfrac{n}{2}$ [2a + (n - 1)d]

=> $S_{7}$ = $\dfrac{7}{2}$ [2a + (7 - 1)d]

=> 182 = $\dfrac{7}{2}$ [2a + 6d]

=> $\dfrac{182\:\times\:2}{7}$ = 2(a + 3d)

=> 52 = 2(a + 3d)

=> 26 = a + 3d _______ (eq 1)

_______________________________

• If its 4th and the 17th terms are in the ratio 1:5.

$a_{n}$ = a + (n - 1)d

» $a_{4}$ = a + (4 - 1)d

» $a_{17}$ = a + (17 - 1)d

• A.T.Q.

=> $\dfrac{a\:+\:(4\:-\:1)d}{a\:+\:(17\:-\:1)d}$ = $\dfrac{1}{5}$

=> $\dfrac{a\:+\:3d}{a\:+\:16}$ = $\dfrac{1}{5}$

Cross-multiply them

=> 5(a + 3d) = a + 16d

=> 5a + 15d = a + 16d

=> 5a - a = 16d - 15d

=> 4a = d

=> d = 4a _____ (eq 2)

• Put value of d in (eq 1)

=> 26 = a + 3(4a)

=> 26 = a + 12a

=> 26 = 13a

=> a = 2

• Put value of a in (eq 2)

=> d = 4(2)

=> d = 8

______________________________

• We have to find AP.

a = 2

a + d = 2 + 8 = 10

a + 2d = 2 + 2(8) = 2 + 16 = 18

_____________________________

AP is 2,10, 18, .....

_________ [ANSWER]