Question

if 6th and 11th terms of an AP are 60 and 80 respectively, find the sum of the first ten terms​

Answer 1

Step-by-step explanation:

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Answer 2

SOLUTION:-

Given:

⚫First term= a

⚫common differenc= d

Then,

${}^{a} n = a + (n - 1)d$

Therefore,

$= > {}^{a} 6 = a + (6 - 1)d \: \: and \: {}^{a} 11 = a + (11 - 1)d \\ \\ = > {}^{a}6 = a + 5d = 60 \: \: and \: \: {}^{a} 11 = a + 10d$

As per given,

a+ 5d = 60...........(1)

a+ 10d= 80..........(2)

Now,

Subtracting equation (1) from (2), we get:

=)5d= 20

=)d= 20/5

=)d= 4

So,

Substituting this value in equation (1), we get:

=) a+ 5(4)= 60

=) a+ 20= 60

=) a= 60-20

=) a= 40

Therefore,

a= 40 & d= 4

Now,

Their sum is:

Using Formula:

Sn= n/2 [2a +(n-1)d]

=)10/2 [2× 40+ (10-1)4]

=) 5[80 + (9)4]

=) 5[ 80+ 36]

=) 5[116]

=) 580 [answer]

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