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

Answer: 18th term of the AP is 0

Step-by-step explanation:

    7(a+6d)=11(a+10d)

=> 7a+42d=11a+110d

=> -4a=68d

=> a=-17d

=> a+17d=0

Since a₁₈=a+17d

The 18th term is 0

Answer 2

${\huge{\star{\underbrace{\tt{\red{Answer}}}}}}{\huge{\star}}$

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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