Question

sum of first 55 terms in an A.P. is 3300,findiys28 term​

Answer 1

Answer:

S55=3300

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

S55=55/2[2a+(55-1) d]

3300=55/2[2a+54d]

3300=55/2×2(a+27d)

3300=55(a+27d)

3300/55=(a+27d)

60=(a+27d)....................(1)

now an=a+(n-1) d

a28=a+(28-1) d

a28=a+27d

a28=60................. (from eqn 1)

Hence 28 term =60

Answer 2

Answer:-

$\boxed{ \bf{t_{28} = 60}}$

Step - by - step explanation :-

To find :-

Find 28th term of the given AP.

Given :-

55th term is 3300.

Solution:-

Let first term of this AP is "a"

Common difference is "d"

According to the question-

$\: \bf{s_{55} = 3300} \\ \\ \bf{ \frac{55}{2} \bigg(2a + (55 - 1)d \bigg) = 3300} \\ \\ \bf{\frac{55}{2} \bigg(2a + 54d \bigg) = 3300} \\ \\ \bf{ 55(a + 27d) = 3300} \\ \\ \bf{a + 27d \: = 60 }\: \: \: ......(1)$

And also ,

We know that,

$\bf{t_{28} \: = a \: + (28 - 1)d} \\ \\ \bf{ t_{28} \: = a + 27d \: } \: ......(2)$

On comparing eq (1) and (2)

We get,

$\boxed{ \red{\bf{t_{28} = 60}}}$

Hope it helps you.