Question

if 7 times the 7th term of an ap is equal to the 11 times the 11th term then what will be its 18th term?​

Answer 1

Answer:

$t _{18} = 0$

Step-by-step explanation:

Given: $7 \times t _{7} = 11 \times t _{11} \: \: - - - - - (i)$

To find:

$t _{18}$

Solution:

As,

$t _{n} \: = a + (n - 1) \times d$

Therefore,

$t _{7} = a + (7 - 1)d \\ t _7 = a + 6d$

Similarly,

$t _{11} \: = a + (11 - 1)d \\ t _{11} = a + 10d$

Put the values in Equation (i) we get,

$7 \times (a + 6d) = 11 \times (a + 10d) \\ 7a + 42d \: = 11a \: + 110d \\ 7a - 11a \: = 110d - 42d \\ - 4a = 68d \\ - a \: = 17d \\ a = - 17d$ -----(ii)

Now,

$t _{18} = a + 17d \\ t _{18} = - 17d \: + 17d \: [as \: a = -17d] \: \\ t _{18} \: = 0$

Therefore, 18th term is 0.

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Answer 2

$\huge{\sf{\underline{\underline{AnSwEr:-}}}}$

$\large\sf\therefore\red{a_{18}=0 }$

$\large{\sf{\underline{\underline{GivEn:-}}}}$

$\bullet$ $\large\tt\ 7a_7= 11a_{11}$

$\large{\sf{\underline{\underline{To \: FiNd :-}}}}$

$\bullet$ $\large\tt\ a_{18} \: term$

$\large{\sf{\underline{\underline{STep \: bY \: STep \: expLanTion :-}}}}$

As we know ;

$\large\green{\boxed{\sf a_n =a+(n-1)d }}$

$\large\sf\ AccoRdinG \: tO \: QueStion:--$

______________________

$\large\star\tt\ 7a_7 = 11a_{11}$

$\large\tt\to\ 7[a+(7-1)d] =11[a+(11-1)d$

$\large\tt\to\ 7[a +6d ] =11[ a +10d ]$

$\large\tt\to\ 7a+ 42d = 11a +110d$

$\large\tt\to\ 42d -110d =11a- 7a \\ \\ \large\tt\to\ -68d = 4a \\ \\ \large\tt\to\ 4a+68d =0 \\ \\ \large\tt\to\ 4(a+17d) =0 \\ \\ \large\tt\to\ a+17d = \frac{0}{4} \\ \\ \large\tt\to\ a+17d =0 ------(1)$

Now,

_____________

$\large\tt\ a_{18} = a+(18-1)d \\ \\ \large\tt\to\ a_{18}= a+ 17d$

By using equation (1):

$\large\tt\ a_{18} = 0$