If the 9^{th} term of an A.P. is 35 and the 1 | vedclass

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

Question

(B) The $n^{th}$ term of an $A.P.$ is given by $T_n = a + (n-1)d$,where $a$ is the first term and $d$ is the common difference.Given $T_9 = a + 8d = 3

Vedclass · Accepted Answer

$79$

Vedclass · Answer

(B) The $n^{th}$ term of an $A.P.$ is given by $T_n = a + (n-1)d$,where $a$ is the first term and $d$ is the common difference.Given $T_9 = a + 8d = 35$ (Equation $1$).Given $T_{19} = a + 18d = 75$ (Equation $2$).Subtracting Equation $1$ from Equation $2$:$(a + 18d) - (a + 8d) = 75 - 35$$10d = 40$$d = 4$.Substituting $d = 4$ into Equation $1$:$a + 8(4) = 35$$a + 32 = 35$$a = 3$.The $20^{th}$ term is $T_{20} = a + 19d$.$T_{20} = 3 + 19(4) = 3 + 76 = 79$.

If the $9^{th}$ term of an $A.P.$ is $35$ and the $19^{th}$ term is $75$,then its $20^{th}$ term will be:

Similar Questions

$0.5737373...... = $

If the sum of the first $6$ terms is $9$ times the sum of the first $3$ terms of the same $G.P.$,then the common ratio of the series will be

The sum of all two-digit numbers which,when divided by $4$,yield unity as a remainder is:

The sum of the series $\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots$ up to infinity is:

Let ${S_n}$ denote the sum of $n$ terms of an $A.P.$ If ${S_{2n}} = 3{S_n}$,then the ratio $\frac{{{S_{3n}}}}{{{S_n}}}$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module