$7 a_7 = 11 a_{11}$

In a non-constant arithmetic progression , 7 times the $7^\text{th}$ term is equal to 11 times the $11^\text{th}$ . Find the $18^\text{th}$ term of this arithmetic progression.

The answer is 0.

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Nice solution (+1) but what if $a_7 = a_{11} = 0$ where $a = d = 0$ . In other words what of it is a constant arithmetic progression? So just for clarity plz mention that it is a non-constant AP @Abhiram Rao . Nice question.

Ashish Menon
- 5 years ago

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Thanks! modified.

Sravanth C.
- 5 years ago

**
It's an Easy A.P
**

$T_{n}= a + (n-1)d$
*
(Formula for finding the
$n^{th}$
term)
*

*
So, By the problem
*

$→7(a + (7-1)d) = 11(a+ (11-1)d) → 7(a + 6d)=11(a + 10d)$

$→ (7a+42d)=(11a + 110d)$

$→ 4a=-68d → 2a=-34d$

$→ a=-17d$

Hence, $T_{18} = -17d +(18 - 1)d)\ ⇶ T_{18} = 17d -17d = \boxed{\boxed{0}}$

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Yeah it's easy :p

Abhiram Rao
- 5 years, 1 month ago

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ATQ.

7(a+6d)=11(a+10d) ,

7a+42d=11a + 110d

, -4a= 68d

, a=-17d
Now

a(18)= a + 17d

, = -17d+17d

, =0

Ans. = 0

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I request you to use latex. Because it will be nice if you use latex for solution.

Samara Simha Reddy
- 5 years, 2 months ago

The $n$ th term of an AP is given by $\boxed{a + (n-1)d}$

Therefore, from the above question, we have:

$\implies 7(a + 6d) = 11(a + 10d) \implies 4a = -68d \implies 4a + 68d = 0 \implies 4(a + 17d) = 0 \implies a + (18-1)d = 0 \implies \boxed{a_{18} = 0}$ , which is the required answer.

I still can't figure out why such an easy question has been rated Level 3 XD

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It confuses some guys. Yeah , but it's an easy one :p

Abhiram Rao
- 5 years, 1 month ago

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Lol, AP is every math geek's favourite :)

Arkajyoti Banerjee
- 5 years, 1 month ago

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$\large \text{The formula for finding the specific term is } a_n = a + (n-1) \times d\\ \large \text{From the above question, We get} 7 \times (a + 6d) = 11 \times (a+10d)\\ \large \implies 7a + 42d = 11a + 110d\\ \large \implies 4a + 68d = 0\\ \large \implies 4 \times (a + 17d) = 0\\ \large \implies 4 \times a_{18} = 0\\ \large \implies a_{18} = \boxed{0}.$