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

"> 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

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

Aptitude Sequences and Series in Aptitude 3 months ago

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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

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Posted on 21 Jun 2025, this text provides information on Aptitude related to Sequences and Series in Aptitude. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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