Question

find the sum of the two middle most terms of the a.p -4/3,-1,-2/3,...........11/3.

Answer 1

Here in the given A.P the first term a , = -4/3
an = 11/3   and d = 1/3

an = a + ( n - 1 ) d
11/3 = -4/3 + ( n - 1 ) 1/3
5 = n/3 - 1/3
5 + 1/3 = n/3
16/3 = n/3
16 = n

Now the middle term = n/2 , n/2 + 1
= 8th term and 9th term.

a + 7d + a + 8d 

2a + 15d
 2(-4/3) + 15(1/3)

-8/3 + 5

-8 + 15
    3 

= 7/3

Answer 2

actually there is a mistake in question

this question is from rd sharma pg .5.26 new class 10 edition

the last term is 4 1/3 which is 13 /3

and the answer given is 3

so the question is wrong

NOTE:

if you want to verify this there is the same question solved by golda . math teacher in brainly itself

here is the solution given by brainly

Solution:-

There is a mistake in this question. The first term should be - 4/3 instead of 

- 4, otherwise the common difference will not be equal.

The AP with the first term is- 4/3 and the last term is 13/3.

The nth term of the AP is given by a + (n - 1)d = Tn. Here a is the first term and d is the common difference.

Tn is the last term.

d = Tn + 1 - Tn, here n is the natural number.

So, d = - 1 - (-4/3)

common difference = 1/3

So, using this formula ----- a + (n - 1)d = Tn  we get

⇒ - 4/3 + (n - 1)1/3 = 13/3

⇒ (-5 +n)/3 = 13/3

3n = 39 + 15

3n = 54

n = 18

So, there are 18 terms in the given AP.

The two middle most terms are 9th term and 10th term.

Hence, T₉ = - 4/3 + (9 - 1)1/3 

T₉ = 4/3

T₁₀ = - 4/3 + (10 - 1)1/3

T₁₀ = 5/3

So, sum of the two middle most terms of the given AP = 4/3 + 5/3

= 9/3

= 3

Answer

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