Question

Sum of first 55 terms of an A.P. is 3300,find its 28th term

Answer 1

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

∴ t28 = a + 27d

∴ t28 = 60 [From (i)]

∴ Twenty eighth term of A.P. is 60.

Answer 2

Hey there, 
Let first term of the AP be $a$ and common difference be $d$ 
Now, we know that Sum of first $n$ terms of an AP is given by

$S_n = \frac{n}{2} \left( 2a+\left( n-1 \right) d \right) \\ \\ Here, \, S_{55} = \frac{55}{2} \left( 2a+\left( 55-1 \right) d \right) \\ \\ \implies 3300 = \frac{55}{2} \left( 2a+54d \right) \\ \\ \implies \frac{3300}{55} = a+27d \\ \\ \implies 2a+27d=60$

Now, $n^{th}$ term of an AP is given as 
$T_n = a+(n-1)d \\ \\ \implies T_{28} = a+27d \\ \\ \implies \boxed{T_{28}=60}$ 

Hope it helps
Purva
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