Question

In an A.P. 19^th term is 52 and 38^th term is 128, find sum of first 56 terms. 

Answer 1

For an A.P., Let a be the first term and d be the common difference.

t₁₉ = 52 and t₃₈ = 128           ...(Given)

Since t_n = a + (n – 1)d

For t₁₉ = 52,

∴ t₁₉ = a + (19 – 1)d

∴ 52 = a + 18d

∴ a + 18d = 52              ...(i)

For t₃₈ = 128,

 t₃₈ = a + (38 – 1)d

∴ 128 = a + 37d

∴ a + 37d = 128           ...(ii)

Subtracting equation (i) from (ii), we get,

\[\begin{array}{l}  
\phantom{\texttt{0}}\texttt{ a + 37d = 128}\\ \phantom{\texttt{}}\texttt{- a + 18d = 52}\\ \hline\phantom{\texttt{}}\texttt{  (-) (-) (-)}\\ \hline \end{array}\]

∴ 19d = 76

∴ d = 4

Substituting d = 4 in equation (i),

a + 18d = 52

∴ a + 18 × 4 = 52

∴ a + 72 = 52

 ∴ a = 52 – 72

 ∴ a = – 20

Now, `S_n = n/2 [ 2a + (n - 1)d]`

∴ S₅₆ = `56/2 [2 × (– 20) + (56 - 1) × 4]`

∴ S₅₆ = 28(– 40 + 220)

∴ S₅₆ = 28 × 180

∴ S₅₆ = 5040

∴ The sum of the first 56 terms is 5040.