Question

in an AP 6th term is 1 more than twice to the 3rd term. The sum of 4th and 5th terms is 5 times the 2nd term. Find 10term of an AP​

Answer 1

Given : In an Arithmetic Progression sixth term is one more than twice the third term. The sum of the fourth and fifth terms is five times the second term.

To find : the tenth term of the Arithmetic Progression.

Solution:

Let say AP

a , a + d , a + 2d ..........

Sixth term = a + 5d

Third term = a + 2d

sixth term is one more than twice the third term.

=> a + 5d = 2(a + 2d) + 1

=> a + 5d = 2a + 4d + 1

=> d - a = 1

=> d = a + 1

sum of the fourth and fifth terms is five times the second term.

fourth term = a + 3d

Fifth term = a + 4d

Second term = a + d

(a + 3d) + (a + 4d) = 5(a + d)

=> 2a + 7d = 5a + 5d

=> 2d = 3a

=> 2(a + 1) = 3a

=> 2a + 2 = 3a

=> a = 2

d = a + 1 = 2 + 1 = 3

AP is

2 , 5 , 8 , 11 ............

10th term = a + 9d = 2 + 9(3) = 29

29 is 10th term of AP

Answer 2

Answer:-

$\sf{The \ 10^{th} \ term \ of \ the \ AP \ is \ 29.}$

Given:

• In an AP 6th term is 1 more than twice to the 3rd term.

• The sum of 4th and 5th terms is 5 times the 2nd term.

To find:

• $\sf{The \ 10^{th} \ term \ of \ the \ AP}$

Solution:

$\boxed{\sf{t_{n}=a+(n-1)d}}$

$\sf{According \ to \ the \ first \ condition.}$

$\sf{t_{6}=(2\times \ t_{3})+1}$

$\sf{a+5d=2(a+2d)+1}$

$\sf{\therefore{a+5d=2a+4d+1}}$

$\sf{\therefore{a-2a+5d-4d=1}}$

$\sf{\therefore{-a+d=1...(1)}}$

$\sf{According \ to \ the \ second \ condition.}$

$\sf{t_{4}+t_{5}=5\times(t_{2})}$

$\sf{\therefore{a+3d+a+4d=5\times(a+d)}}$

$\sf{\therefore{2a+7d=5a+5d}}$

$\sf{\therefore{2a-5a+7d-5d=0}}$

$\sf{\therefore{-3a+2d=0...(2)}}$

$\sf{Multiply \ equation (1) \ by \ 2, \ we \ get}$

$\sf{-2a+2d=2...(3)}$

$\sf{Subtract \ equation (3) \ from \ equation (2)}$

$\sf{-3a+2d=0}$

$\sf{-}$

$\sf{-2a+2d=2}$

___________________

$\sf{-a=-2}$

$\boxed{\sf{\therefore{a=2}}}$

$\sf{Substitute \ a=2 \ in \ equation (1), \ we \ get}$

$\sf{-2+d=1}$

$\sf{\therefore{d=1+2}}$

$\boxed{\sf{\therefore{d=3}}}$

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$\sf{Here, \ a=2 \ and \ d=3}$

$\sf{t_{10}=a+(10-1)d}$

$\sf{\therefore{t_{10}=a+9d}}$

$\sf{\therefore{t_{10}=2+9(3)}}$

$\sf{\therefore{t_{10}=2+27}}$

$\sf{\therefore{t_{10}=29}}$

$\sf\purple{\tt{\therefore{The \ 10^{th} \ term \ of \ the \ AP \ is \ 29.}}}$