Question

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

Answer 1

Given:-

• The 15th term of an AP is 3 more than twice its 7th term
• 10th term of the AP is 41

To Find:-

• nth term of AP

Step-by-step explanation:

We know the formula of nth term of AP,

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

where,

• a = 1st term
• d = common difference

Therefore,

15th term of AP is,

${ \sf{T_{15} = a + (15 - 1)d}}$

$\implies \: { \sf{T_{15} = a + 14d}}$

And the 7th term of AP is,

${ \sf{T_{7} = a + (7- 1)d}}$

$\implies \: { \sf{T_{7} = a + 6d}}$

According to the given question,

$\implies \sf \: T_{15} = 2T_{7} + 3$

$\implies \sf a + 14d = 2(a + 6d) + 3$

$\implies \sf a + 14d = 2a + 12d + 3$

$\implies \sf a = 2d + 3$

Now, 10th term of AP is,

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

$\implies \: { \sf{T_{10} = a + 9d}}$

According to the given condition,

$\implies \sf \: a + 9d = 41....(i)$

Putting the value of a,

$\implies \sf \: 2d + 3 + 9d = 41$

$\implies \sf \: 11d = 41 - 3 = 38$

$\implies \sf \: d = \dfrac{38}{11} = 3.45$

Putting the value of d in equation (i)

$\implies \sf \: a + 9 \times 3.45 = 41$

$\implies \sf \: a = 41 - 31.09 = 9.91$

nth term of AP,

$\implies \sf T_{n} = 9.91 + (n - 1) \times 3.45$

$\implies \sf T_{n} = 9.91 + 3.45n - 3.45$

$\implies \sf T_{n} = 6.46 + 3.45n$

Answer 2

$\sf {\blue {Given : }}$

• 15th term of an AP is 3 more than twice its 7th term .
• 10th term of the AP is 41 .

$\sf {\purple {To\: find : }}$

• nth term of the AP

$\sf {\red {Solution : }}$

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

$\sf : \implies 15th \:term\: of \:an \:AP \:is\:3 \:more \:than \:twice\: its\: 7th\: term .$

15 term = a + (15-1) d = a + 14 d

7th term = a + (7-1) d = a + 6d

So , according to given data ..

$: \implies \sf a + 14d = 2a + 12d + 3$

$:\implies \sf a = 2d + 3$

Now, 10th term of AP is,

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

$:\implies \: { \sf{T_{10} = a + 9d}}$

According to the given condition,

$:\implies \sf \: a + 9d = 41....(1)$

Substituting the value of a,

$: \implies \sf \: 2d + 3 + 9d = 41$

$:\implies \sf \: 11d = 41 - 3 = 38$

$: \implies \sf \: d = \dfrac{38}{11}$

Substituting the value of d in equation (1)

$: \implies \sf \: a + 9 \times 3.45 = 41$

$: \implies \sf \: a = 41 - 31.09 = 9.91$

nth term of AP,

$:\implies \sf T_{n} = 9.91 + (n - 1) \times 3.45$

$:\implies \sf T_{n} = 9.91 + 3.45n - 3.45$

${\sf{\orange {Answer : }}}$

${\boxed {\underline {\sf {T_{n} = 6.46 + 3.45n}}}}$