हिंदी

A man borrows Rs 8000 and agrees to repay with a total interest of Rs 1360 in 12 monthly installments. Each installment being less than the preceding one by Rs 40. - Algebra

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प्रश्न

A man borrows Rs 8000 and agrees to repay with a total interest of Rs 1360 in 12 monthly installments. Each installment being less than the preceding one by Rs 40. Find the amount of the first and last installments.

योग

उत्तर

Money he borrows = Rs. 8000

Total interest = Rs. 1360 

Total money he will pay after 12 months = Rs (8000 + 1360) = Rs 9360
n = 12 and S12 = 9360

Each installment being less than the preceding one by Rs 40.
d = −40
Now,

Sn = `n/2 [2a + (n − 1)d]`

∴ S12 = `12/2 [2a + (12 − 1)(− 40)]`

∴ 9360 = 6[2a + 11 × (− 40)]

∴ 9360 = 6(2a − 440)

∴ `9360/6 = 2a − 440`

∴ 1560 = 2a − 440

∴ 1560 + 440 = 2a

∴ 2000 = 2a

∴ a = `2000/2`

∴ a = 1000

tn = a + (n − 1)d

∴ t12 = 1000 + (12 − 1)(− 40)

∴ t12 = 1000 + 11(− 40)

∴ t12 =  1000 − 440

∴ t12 = 560

∴ Amount of the first instalment is Rs. 1000 and that of the last instalment is Rs. 560.

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अध्याय 3: Arithmetic Progression - Practice Set 3.4 [पृष्ठ ७८]

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बालभारती Algebra (Mathematics 1) [English] 10 Standard SSC Maharashtra State Board
अध्याय 3 Arithmetic Progression
Practice Set 3.4 | Q 2 | पृष्ठ ७८

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