Which term of the A.P. 5, 10, 15, ldots excee | vedclass

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

Question

(D) Given $A.P.$ is $5, 10, 15, \ldots$Here,first term $a = 5$ and common difference $d = 10 - 5 = 5$.The $n^{th}$ term of an $A.P.$ is given by $a_n

Vedclass · Accepted Answer

$57$

Vedclass · Answer

(D) Given $A.P.$ is $5, 10, 15, \ldots$Here,first term $a = 5$ and common difference $d = 10 - 5 = 5$.The $n^{th}$ term of an $A.P.$ is given by $a_n = a + (n - 1)d$.The $31^{st}$ term is $a_{31} = 5 + (31 - 1)5 = 5 + 30 	imes 5 = 5 + 150 = 155$.Let the $n^{th}$ term exceed the $31^{st}$ term by $130$,so $a_n = a_{31} + 130$.$a_n = 155 + 130 = 285$.Using the formula $a_n = a + (n - 1)d$:$285 = 5 + (n - 1)5$.$280 = (n - 1)5$.$n - 1 = 280 / 5 = 56$.$n = 56 + 1 = 57$.Therefore,the $57^{th}$ term exceeds the $31^{st}$ term by $130$.

Part $I$	Part $II$
$1$. Sixth term of the $A.P.$ $1, 3, 5, 7, \ldots$	$a$. $105$
$2$. Eleventh term of the $A.P.$ $3, 6, 9, 12, \ldots$	$b$. $11$
$3$. Sixteenth term of the $A.P.$ $4, 6, 8, 10, \ldots$	$c$. $33$
$4$. Twenty-first term of the $A.P.$ $5, 10, 15, 20, \ldots$	$d$. $34$

Which term of the $A.P.$ $5, 10, 15, \ldots$ exceeds its $31^{st}$ term by $130$ (in $^{th}$)?

Similar Questions

For the $AP: -3, -7, -11, \ldots,$ can we find $a_{30} - a_{20}$ directly without actually finding $a_{30}$ and $a_{20}$? Give reasons for your answer.

Find the $100^{th}$ term of the $A.P.$ $50, 56, 62, 68, \ldots$

The $21^{\text{st}}$ term of the $AP$ whose first two terms are $-3$ and $4$ is

For each of the following $A.P.s$,find the $n^{th}$ term: $\frac{4}{3}, 2, \frac{8}{3}, \frac{10}{3}, \ldots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module