Question

if 4 time the 4th terms of an ap is equal to 18 time the 18th term, then find the 22nd term​

Answer 1

GIVEN :-

• 4 times the 4th term of an A.P is equal to 18 times the 18th term.

TO FIND :-

• 22nd term of A.P.

SOLUTION :-

For 'n'th term,

★ a{n} = a + (n - 1)d

• a{n} is 'n'th term.
• a is first term.
• n is total number of terms.
• d is common difference.

So,

♦ 4th term of A.P,

→ a{4} = a + (4 - 1)d

→ a{4} = a + 3d -----(i)

♦ 18th term of A.P,

→ a{18} = a + (18 - 1)d

→ a{18} = a + 17d -----(ii)

♠ According to question,

4 times the 4th terms of an A.P is equal to 18 times the 18th term.

→ 4 × a{4} = 18 × a{18}

Putting values from equation (i) and (ii),

→ 4(a + 3d) = 18(a + 17d)

→ 4a + 12d = 18a + 306d

→ 4a - 18a + 12d - 306d = 0

→ -14a - 294d = 0

Dividing whole equation by-14,

→ a + 21d = 0 ------(iii)

Now , for 22nd term,

→ a{22} = a + (22 - 1)d

→ a{22} = a + 21d

From equation (iii) , a + 21d = 0

→ a{22} = 0

Hence , 22nd term of the A.P is 0.