SAVE P in the first year.
Then, the given SEQUENCE is P + (P + 2000) + (P + 4000) + ..,
which is an AP with a = P , d = (P + 2000) - P = 2000,
n = 10, Sn = 145000

Sn = n/2[2a + (n - 1)d]
145000 = 10/2 [2 x P + 9 x 2000] = 5 [2P + 18000]
2P + 18000 = 145000/5 = 29000
2P = 29000 - 18000
2P = 11000
P = 5500
So, the man save 5500 in the first year .

"> SAVE P in the first year.
Then, the given SEQUENCE is P + (P + 2000) + (P + 4000) + ..,
which is an AP with a = P , d = (P + 2000) - P = 2000,
n = 10, Sn = 145000

Sn = n/2[2a + (n - 1)d]
145000 = 10/2 [2 x P + 9 x 2000] = 5 [2P + 18000]
2P + 18000 = 145000/5 = 29000
2P = 29000 - 18000
2P = 11000
P = 5500
So, the man save 5500 in the first year .

">
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A man saves 145000 in ten years. In each year after the first year. he saved 2000 more than he did in the proceeding year. How much did he save in the first year ?

Aptitude Sequences and Series in Aptitude 3 months ago

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Let the man first SAVE P in the first year.
Then, the given SEQUENCE is P + (P + 2000) + (P + 4000) + ..,
which is an AP with a = P , d = (P + 2000) - P = 2000,
n = 10, Sn = 145000

Sn = n/2[2a + (n - 1)d]
145000 = 10/2 [2 x P + 9 x 2000] = 5 [2P + 18000]
2P + 18000 = 145000/5 = 29000
2P = 29000 - 18000
2P = 11000
P = 5500
So, the man save 5500 in the first year .

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Posted on 20 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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