Question

In an AP.5th term is 17 and 9th term exceeds 2nd term by 35. Find the AP.

Answer 1

HELLO DEAR,

let first term a and common difference be d.

given that:- a₅ = 17
⇒a + 4d = 17-----------( 1 )

and 9th term exceeds 2nd term by 35.
a₉ = 35 + a₂
⇒a + 8d = 35 + a + d
⇒8d - d = 35
⇒7d = 35
⇒d = 5 [put in --------( 1 )]

we get,

a + 4(5) = 17
⇒a = 17 - 20
⇒a = - 3

hence, a = -3 , d = 5
NOW, ap: -3 , (-3 + 5) , (-3 + 2*5) , (-3 + 3*5) , ........
ap: -3 , 2 , 7 , 12........


I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Let , the first term be 'a' and the common difference be 'd'

Given,

$t_{5}$ = 17
=> a + 4d = 17 ----------(i)

Again,

$t_{9}$ = $t_{2}$ + 35
=> a + 8d = a + d + 35
=> 8d - d = 35
=>7d = 35
=> d = $\frac{35}{7}$
=> d = 5


From (i)

a + 4d = 17
=> a + 4 × 5 = 17
=>a = 17 - 20
=> a = -3

Required AP

$t_{1}$ = -3

$t_{2}$ = a + d
= -3 +5
=2

$t_{3}$ = a + 2d
= -3 + 2×5
=-3 + 10
=7

$t_{4}$ = a + 3d
= -3 + 3 × 5
= -3 + 15
=12