How About The $(7+11=18)^\text{th}$ Term?

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

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

Alternatively, we could consider the sub-progression starting at the 7th term. Then, $7a=11 (a+4p)$ , and thus $a=-11p$ . The 18th term is then $a+11p=-11p+11p=0.$

We know a(n)= a + (n-1)d
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

If pTp=qTq then a(p+q)=0

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

The sequence can be expressed in the form Tn = mn + c. Substituting n = 7 and 11 into the equation we get: 49m + 7c = 121m + 11c We simplify this into c = -18m Substituting that into the original equation: Tn = m(n-18) Substituting 18 into it, we get 0.