Question

The 15th term of an AP is 3 is more then twice its 7th term if the 10th term of the AP is 41 then find its nth term.

Answer 1

a+14d=2(a+6d)+3
⇒a-2d+3=0 ...(i)
again
a+9d=41 (ii)

subtracting i from ii

11d=44
⇒d=4
hence from ii
a=5
hence nth term = 5+ (n-1) x4
=4n+1

Answer 2

The nth term of the AP is $t_n=1 + 4n$.

• The  nth term of an AP is given by the formula $t_n=a + (n-1)d$ , where a is the first term ,d is the common difference.

• Therefore , 15th term of an AP is 3 more than twice the 7th term of the AP

$t_{15} = 2t_{7} + 3$

$a + (15 - 1)d = 2[a + (7-1)d] + 3$

$a + 14d = 2a+12d + 3$

$a + 14d = 2a+12d + 3$

a - 2d = -3     (Equation 1)

• Now,

10th term of the AP is 41

$t_{10}=41$

a + 9d = 41    (Equation 2)

• On subtracting equation 2 from equation 1 , we get

d = 4

• Now substituting the value of d in equation 1, we get

a = 5

Therefore, the nth term of the AP becomes

$t_n=5 + (n-1)4$

$t_n=5 + 4n - 4$

$t_n=1 + 4n$

• The nth term of the AP is $t_n=1 + 4n$.