Question

If 7 times the 7th term of an A.P. is equal to 11 times its 11th term, then find its 18th term.​

Answer 1

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Given :

7 times the 7th term = 11 times 11th term

To find :

18th term of AP

Solution :

We know that ,

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

${\sf \leadsto \ T_{7} \ = \ a \ + \ (7-1)d}$

${\sf \leadsto \ T_{7} \ = \ a \ + \ 6d}$

Hence 7th term = a+6d

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Similarly,

${\sf \leadsto \ T_{11} \ = \ a \ + \ (11-1)d}$

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

Hence 11th term is a+10d

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Now, According to question

✿ 7 times the 7th term = 11 times 11th term

${\sf{\leadsto \ 7 \times (a + 6d) = 11 \times (a + 10d)}}$

${\sf{\leadsto \ 7a + 42d \ = \ 11a + 110d}}$

${\sf{\leadsto \ 7a - 11a \ = \ 110d - 42d}}$

${\sf{\leadsto \ -4a \ = \ 68d }}$

${\sf{\leadsto \ a \ = \ \frac{ 68d}{-4} }}$

${\sf{\leadsto \ a \ = \ -17d}}$...(i)

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Now ,

${\sf \leadsto \ T_{18} \ = \ a \ + \ (18-1)d}$

Substituting value of a from (i)

${\sf \leadsto \ T_{18} \ = \ -17d \ + \ 17d}$

${\sf \leadsto \ T_{18} \ = \ 0}$

Hence the 18th term of AP is 0.

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Answer 2

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Given :

• 4 times the fourth term of an A.P. is equal to 7 times the 7th term

To find :

• 11th term

Solution :

According to the question:

4(a+3d) = 7(a+6d)

4a+12d=7a+42d

-3a+12d=42d

-3a-30d = 0

3a + 30d = 0

3(a+10d) = 0

a+10d = 0

$\sf{a_{11}}= 0$

Hence, the 11th term is 0