A man saved ₹ 16500/- in ten years . in each year after the first he

1.	A man saved ₹ 16500/- in ten years . in each year after the first he saved ₹ 100/- more then he did in the preceding year. How much did he saved in the first year?
Answer» Given : A man saved 16500/- in ten years To find : His saving in the first year Let he saved Rs. x in the first year Since each year after the first he saved 100/- more then he did in the preceding year So, A.P will be x, 100 + x, 200 + x……………….. Where x is first term and common difference, d = 100 + x – x = 100 We know, S_n is the sum of n terms of an A.P Formula used : s_n = \(\frac{n}{2}\){2a + (n-1)d} Where a is first term, d is common difference and n is number of terms in an A.P. According to the question : S_n = 16500 and n = 10 Therefore, s₁₀ = \(\frac{10}{2}\){2x + (10-1)100} ⇒ 16500 = 5{2x + 9(100)} ⇒ 16500 = 5(2x + 900) ⇒ 16500 = 10x + 4500 ⇒ -10x = 4500 – 16500 ⇒ –10x = –12000 ⇒ x = \(\frac{-12000}{-10}\) ⇒ x = 1200 Hence, his saving in first year is 1200.

A man saved ₹ 16500/- in ten years . in each year after the first he saved ₹ 100/- more then he did in the preceding year. How much did he saved in the first year?

Answer»

Given : A man saved 16500/- in ten years

To find : His saving in the first year

Let he saved Rs. x in the first year

Since each year after the first he saved 100/- more then he did in the preceding year

So,

A.P will be x, 100 + x, 200 + x………………..

Where x is first term and

common difference,

d = 100 + x – x = 100

We know,

S_n is the sum of n terms of an A.P

Formula used :

s_n = \(\frac{n}{2}\){2a + (n-1)d}

Where a is first term, d is common difference and n is number of terms in an A.P.

According to the question :

S_n = 16500 and n = 10

Therefore,

s₁₀ = \(\frac{10}{2}\){2x + (10-1)100}

⇒ 16500 = 5{2x + 9(100)}

⇒ 16500 = 5(2x + 900)

⇒ 16500 = 10x + 4500

⇒ -10x = 4500 – 16500

⇒ –10x = –12000

⇒ x = \(\frac{-12000}{-10}\)

⇒ x = 1200

Hence, his saving in first year is 1200.