Question

The 4th term of an A.P. is zero. Prove that the 25th term of the A.P. is three times its 11th term.​

Answer 1

Step-by-step explanation:

Given :-

• The 4th term of an A.P. is zero.

To prove :-

• The 25th term of the A.P. is 3 times its 11th term.

Solution :-

Formula used :

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

• a = 1st term
• d = common difference

Now find the 4th term of the A.P.

★$\sf{T_4=a+(4-1)d}$

$\to\sf{T_4=a+3d}$

According to the question,

a + 3d = 0

→ a = -3d............(i)

Now find the 25th term and 11th term of the A.P.

★ $\sf{T_{25}=a+(25-1)d}$

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

And,

★ $\sf{T_{11}=a+(11-1)d}$

$\to\sf{T_{11}=a+10d}$

Then,

${\sf{25th\:term=3(11th\:term)}}$

→ a+24d = 3(a+10d)

• Put a = -3d from eq(i).

→ -3d + 24d = 3(-3d+10d)

→ 21d = 3 × 7d

→ 21d = 21d

Hence proved that the 25th term of the A.P. is 3 times its 11th term.