Question

the 7th term of an AP is 40. the sum of the first 13th term is​

Answer 1

AnswEr :

• Sum of the first 13th term is 520.

Given :

• 7th term = 40

To Find :

• Sum of the first 13th term = ?

Solution :

Let us assume that the first term be a and the common difference be d.

We know that,

$\sf \dashrightarrow \ \ \ a_n=a+(n-1)d$

$\sf \dashrightarrow \ \ \ a_7=40$

$\sf \dashrightarrow \ \ \ a + ( 7 - 1 )d = 40$

$\sf \dashrightarrow \ \ \ a + 6d = 40$

Now,

$\sf \dashrightarrow \ \ \ S_n=\dfrac{n}{2}(2a+(n-1)d)$

$\sf \dashrightarrow \ \ \ S_{13}=\dfrac{13}{2}(2a+(13-1)$

$\sf \dashrightarrow \ \ \ \dfrac{13}{2}(2a+12d)$

$\sf \dashrightarrow \ \ \ \dfrac{13}{2}\times2(a+6d)$

$\sf \dashrightarrow \ \ \ 13\times40$

$\sf \dashrightarrow \ \ \ { \underline{ \boxed{ \sf{ \red{520}}}}} \ \bigstar$

Therefore, sum of the first 13th term is 520.

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★ Additional Information :

• The general form of an AP is

• a, a + d, a + 2d, a + 3d ....

• Formula to find the nth term of an AP is

• Tn = a + (n – 1) d

where,

1. tn = nth term
2. a= the first term
3. d= common difference
4. n = number of terms

• Formula to find the numbers of term of an AP is

• n= [ (l-a) / d ] + 1

where,

1. n = number of terms
2. a= the first term
3. l = last term
4. d= common difference

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Answer 2

Solution :

Firstly, we know that formula an A.P;

$\boxed{\bf{a_n=a+(n-1)d}}}$

Where as;

• a is the first term.
• d is the common difference.
• n is the term of an A.P.

A/q

$\longrightarrow\sf{a_7=40}\\\\\longrightarrow\sf{a+(n-1)d=40}\\\\\longrightarrow\sf{a+(7-1)d=40}\\\\\longrightarrow\bf{a+6d=40................(1)}$

Now, using formula of the sum of an A.P;

$\boxed{\bf{S_n=\frac{n}{2} \bigg[2a+(n-1)d\bigg]}}}$

$\longrightarrow\sf{S_{13} = \dfrac{13}{2} \bigg[2a+(13-1)d\bigg]}\\\\\\\longrightarrow\sf{S_{13} = \dfrac{13}{2} \bigg[2a+12d\bigg]}\\\\\\\longrightarrow\sf{S_{13} = \dfrac{13}{\cancel{2}} \times \cancel{2} (a+6d)}\\\\\longrightarrow\sf{S_{13} = 13\times (a+6d)}\\\\\longrightarrow\sf{S_{13} = 13\times 40\:\:[from\:eq.(1)]}\\\\\longrightarrow\bf{S_{13} = 520}$

Thus;

The sum of first 13th term is 520 .