Question

Solve previous year Question of iit jee

Chapter :- sequence and series​

Answer 1

Answer:

His total saving from the start of service will be 11040 after 21 months, option (c) is correct

Step-by-step explanation:

Let the time taken to save ₹ 11040 be (n+3) months .

for the first three months , he saves ₹ 200 each month . in (n+3)months

After 3 months , first term "a"will be taken as 240 and common difference "d" as 40 according to given in question ,

total Sum of saving of man

• 3×200 + n/2{ 2× (a )+ (n-1)×d} = 11040

• → 600 + n/2 { 2× 240 + (n-1)40 } = 11040

• → 600 + 20n (n+11) = 11040

• → 30 + n² + 11n = 552

• → n² + 11n - 522

• → n² + 29n - 18n - 522

• → n(n+29) - 18(n+29) = 0

• → n = 18 , n = -29 neglecting

∴ n = 18 , total time = (n+3) = 21 months