Question

ICBSE 2014
44. The sum of 5th and 9th terms of an A.P.is 30. If its 25th term is three times its 8th
term, find
the AP
ICBSE 2014
11​

Answer 1

Answer:

d= -3 and

a= 3

its AP is 3,0,-3etc

Answer 2

Answer:

The A.P. is 3,5,7,9..........

Explanation:

Given :-

• The sum of 5th term and 9th terms of an A.P is 30.
• Its 25th term is 3 times of it's 8th term.

To find :-

• The A.P.

Solution :-

Let the A.P is → a, a+d, a+2d, a+3d

• 1st term = a
• Common difference = d

Formula used :-

${\boxed{\sf{T_n=a+(n-1)d}}}$

★ $\sf{T_5=a+(5-1)d}$

$\to\sf{T_5=a+4d}$

★ $\sf{T_9=a+(9-1)d}$

$\to\sf{T_9=a+8d}$

According to the 1st condition :-

• The sum of 5th term and 9th terms of an A.P is 30.

$\to\sf{T_5+T_9=30}$

$\to\sf{a+4d+a+8d=30}$

$\to\sf{2a+12d=30...........(I)}$

Now find the 25th term and 8th term of the A.P.

★$\sf{T_{25}=a+(25-1)d}$

$\to\sf{T_{25}=a+24d}$

★ $\sf{T_8=a+(8-1)d}$

$\to\sf{T_8=a+7d}$

According to the 2nd condition :-

• Its 25th term is 3 times of it's 8th term.

$\to\sf{T_{25}=3T_8}$

$\to\sf{a+24d=3\times(a+7d)}$

$\to\sf{a+24d=3a+21d}$

$\to\sf{a-3a=21d-24d}$

$\to\sf{-2a=-3d}$

$\to\sf{2a=3d}$

$\to\sf{2a-3d=0...............(ii)}$

Now subtract eq(i) from eq (ii).

2a - 3d -(2a +12d)= 0-30

→ 2a-3d-2a-12d = -30

→ -15d = -30

→ d = 2

• Common difference = 2

Now put d = 2 in eq(ii).

2a - 3d = 0

→ 2a - 3×2 = 0

→ 2a -6 = 0

→ 2a = 6

→ a = 3

Then,

• a = 3
• a+d = 3+2 = 5
• a+2d = 3+2×2 = 7
• a+3d = 3+3×2 = 9

Therefore the A.P. is 3,5,7,9..........