Class 10

Math

All topics

Arithmetic Progressions

The sum of the $5_{th}$ and the $7_{th}$ terms of an AP is 52 and the $10_{th}$ term is 46

then the AP is 1, 6, 11, 16.......

If true then enter $1$ and if false then enter $0$

- 1

Then,

$a_{5}+a_{7}=52$

$a_{10}=46$

$a+4d+a+6d=52$ and

$a+9d=46$

$⇒2a+10d=52....(1)$

$a+9d=46....(2)$

multiplying (2) equation by 2 and we get,

$2a+18d=92....(3)$

now subtract (1) from (3) then we get,

$8d=40$

$⇒d=5$

now put d in (1) equation we get

$a=1$

There for the given series is

$a,a+d,a+2d,a+3d...$

i.e. $1,6,11,16...$