Math, 18.10.2019 11:00, dnyandeepnile1999

The 7th term of an ap is 40 the sum of its first 13th term is a)5oo. b)51o. c)52o. d)53o

Answers: 1

Get

Answers

The correct answer was given: jack626259

$7th \: \: term \: \: = a + 6d = 40 \\ \\ sum \: \: of \: \: 13 \: \: terms \: = \frac{13}{2} (2a + 12d) \\ \\ = \frac{13}{2} (2)(a + 6d) = 13(40) = 520$

The correct answer was given: siya1472006

520

Step-by-step explanation:

40=a+6d

a=40-6d

S13=13/2(2(40-6d)+12d)

S13=13/2(80-0)

S13= 13/2(80)

=520

The correct answer was given: malikfaiz24

Let the first term be a and common difference is d,

a+(7−1)d=40

a+6d=40

Now sum of 13 term is

13/2(2a + 12d)= 13/2*2*40=520

The correct answer was given: jesuiskhushi

okk

bye

$\alpha \beta e \csc( \sec( \cot( \tan( \cos( ln( log( log_{\% \binom{e}{?} }(?) ) ) ) ) ) ) )$

byyyyy

The correct answer was given: Sowdamini

Step-by-step explanation:

This is ur answer dude hope u like it

7th term of an ap is 40. the sum of its first 13th terms is

The correct answer was given: shad007

• Let the first term of an A.P. be a and common difference is d. So the 7th term of the A.P. is 40=> a+6d =40.

Sum of first 13 terms = 13/2(a +(a+12d))= 13/2(2a+12d) =

13/2(2)(a+6d) =13(a+6d) =13(40)=

520 .

The correct answer was given: eka85

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.

⠀⠀━━━━━━━━━━━━★ 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,

tn = nth terma= the first term d= common differencen = number of terms

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

n= [ (l-a) / d ] + 1

where,

n = number of termsa= the first terml = last termd= common difference⠀⠀━━━━━━━━━━━━

The correct answer was given: kalpana8955

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;

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

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