Question

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

Answer 1

Given: The series:  1 + 4 + 6 + 5 + 11 + 6 + ....

To find: The sum to 200 terms of the series.

Solution:

• Now we have given a series: 1 + 4 + 6 + 5 + 11 + 6 + ....

• In the given series, we can observe that that there are two AP in them.

• First AP is (1 + 6 + 11 +…)  

• Second AP is (4 + 5 + 6 +…)  

• Sum of AP = n/2 x (2a + (n-1)d)

• Now using sum of AP formula, we get:

                 S = (1 + 6 + 11 +…  100th term) + (4 + 5 + 6 +  ......100th term)

                 S = 100 { (2 x 1) + (100-1) x 5 } / 2 + 100 { (2 x 4) + (100-1) x 1 } / 2  

                 S = 100 { 2 + (99) x 5 } / 2 + 100 { 8 + (99) x 1 } / 2  

                 S = 50 { 497 + 107 }

                 S = 50 { 604 }

                 S = 30200

Answer:

            So the sum of the given series is 30200.

Answer 2

Answer:

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