Find the 12^{text{th}} term from the end of t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $AP: -2, -4, -6, \ldots, -100$.Here,the first term $(a) = -2$,the common difference $(d) = -4 - (-2) = -2$,and the last term $(l) = -100$.We

Vedclass · Accepted Answer

$-78$

Vedclass · Answer

(A) Given $AP: -2, -4, -6, \ldots, -100$.Here,the first term $(a) = -2$,the common difference $(d) = -4 - (-2) = -2$,and the last term $(l) = -100$.We know that the $n^{	ext{th}}$ term of an $AP$ from the end is given by the formula $a_n = l - (n - 1)d$,where $l$ is the last term and $d$ is the common difference.Therefore,the $12^{	ext{th}}$ term from the end is:$a_{12} = -100 - (12 - 1)(-2)$$a_{12} = -100 - (11)(-2)$$a_{12} = -100 + 22$$a_{12} = -78$.Hence,the $12^{	ext{th}}$ term from the end is $-78$.

Column $A$	Column $B$
$(A_{1}) \quad 2, -2, -6, -10, \ldots$	$(B_{1}) \quad \frac{2}{3}$
$(A_{2}) \quad a = -18, n = 10, a_{n} = 0$	$(B_{2}) \quad -5$
$(A_{3}) \quad a = 0, a_{10} = 6$	$(B_{3}) \quad 4$
$(A_{4}) \quad a_{2} = 13, a_{4} = 3$	$(B_{4}) \quad -4$
	$(B_{5}) \quad 2$
	$(B_{6}) \quad \frac{1}{2}$
	$(B_{7}) \quad 5$

Find the $12^{\text{th}}$ term from the end of the $AP: -2, -4, -6, \ldots, -100$.

Similar Questions

For the finite $A.P.$ $40, 35, 30, \ldots, -200,$ find the $10^{th}$ term from the end.

Can any term of the $A.P.$ $14, 18, 22, \ldots$ be $142$? If yes,which term?

Which of the following is not an $A.P.$?

The first term of a finite $A.P.$ is $5$ and its last term is $45$. If the sum of all the terms is $500$,there are $\ldots$ terms in the $A.P.$

Vedclass Test Series

Exam Paper Generator

Online Exam Module