Question

the 7th and 17th term of an arithmetic progression are 33 and 83 respectively find the ap


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Answer 1

$\textbf{Concept:}$

$\text{The n th term of the A.P a, a+d, a+2d,...... is }$

$\boxed{\bf\,t_n=a+(n-1)d}$

$\textbf{Given:}$

$t_7=33\;\text{and}\;t_{17}=83$

$\text{we get,}$

$a+6d=33$........(1)

$a+16d=83$.......(2)

$\text{subtract (2) from (1) }$

$-10d=-50$

$\implies\boxed{d=5}$

$\text{put d=5 in (1)}$

$a+30=33$

$\implies\boxed{a=3}$

$\therefore\textbf{The required A.P is 3, 8, 13........}$

Find more:

If 8 times the 8th term of an AP is equal to 15 times its 15th term then find the 23rd term.

https://brainly.in/question/6155736#

Answer 2

The ap is 3,8,13,18,23,...

• Let the first term of the arithmetic progression be a and the common difference be d
• Now it is given that the 7th and 17th term of the arithmetic progression are 33 and 83 respectively
• Now applying arithmetic progression formula we get a+(7-1)d = 33 and a+(17-1)d = 83
• Or a+6d=33 and a+16d=83
• Now subtracting first equation from second equation we get 10d = 83-33 = 50, Or d = 5
• putting the value of d in first equation we get a = 33-6*5 = 33-30 = 3
• So the arithmetic progression is a,a+d,a+2d,a+3d,a+4d,... which is equal to 3,8,13,18,23,...