Question

the 15th term of a AP is 3 more than twice its 7th term .If the tenth term of the AP is 41 , find its 31st term

Answer 1

Step-by-step explanation:

$\Huge\red{\mathfrak{ Answer }}$

Let initial term be a

and common difference be d

then 15th term = a+ 15d

7th term= a+7d

given

a+15d= 2(a+7d)+3

=> a+15d= 2a+14d+3

=> 2a-a+3= 15d-14d

=> a+3= d

So

10th term= a+10d= 41

=> a+ 10(a+3)= 41

=> a+10a+30= 40

=> 11a= 10

=>a = 10/11

then. d = a+3= 10/11+3= 10+33/11= 43/11

So

31st term = a+31d

= 10/11+ 31(43/11)

= 10/11+31*43/11

=10/11+1333/11

= 10+1333/11

= 1343/11

hope it helps ✌