Find the sum to 200 terms of the series. 1 + 4 + 6 + 5 + 11 + 6 + ...?

Here, the 1st AP is (1 + 6 + 11 + ...)
and 2nd AP is (4 + 5 + 6 + ...)
1st AP = (1 + 6 + 11 + ..)
Here , common difference = 5 and the number of terms = 100
∴ Sum of series = S₁ = n/2[2a + (n - 1)d]
= 100/2 [2 x 1 + (100 - 1) x 5 ]
= 50[2 + 99 x 5 ]
= 50 x 497
= 24850
2nd AP = (4 +5 + 6 + ...)
Here, common difference = 1
and number of terms = 100
S₂ = 100/2[2 x 4 + 99 x 1]
= 50 x 107 = 5350
∴ Sum of the given series
= S₁ + S₂ = 24850 + 5350
= 30200