Question

sum of first 55 terms in an A.P. is 3300, find its28th term

Answer 1

 The sum of the first 55 terms of an A. P. is 3300. Find the 28th term.


Sol. S55 = 3300 [Given]


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


∴ a + 27d= 60 ......(i)



Now, tn = a + (n – 1) d



∴  t28 = a + (28 – 1) d





HOPE IT HELP

∴ t28 = a + 27d


∴ t28 = 60 [From (i)]


∴ Twenty eighth term of A.P. is 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.