Math, asked by yashwanthirandhi, 5 days ago

please answer these questions

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Answered by tennetiraj86
3

Answer:

i) (1)

ii)(2)

iii)(3)

iv)(1)

Step-by-step explanation:

Solution :-

Total amount repaid by Saurabh = Rs. 590000

The starting ins.talment = Rs. 5000

Amount is increased in every month in the ins.talment = Rs. 500

This situation is related to Arithmetic Progression.

We have ,

First term (a) = Rs. 5000

Common difference (d) = Rs. 500

The second ins.talment = 5000+500 =Rs. 5500

The AP: 5000,5500,6000,...,590000

We know that

nth term of an AP = an = a+(n-1)d.

i)The amount is paid by him in 20th ins.talment

= 20th term of the AP

=> a 20 = a+(20-1)d

=> a 20 = a+19d

=> a 20 = 5000+(19×500)

=> a 20 = 5000+9500

=> a 20 = 14500

The 20th ins.talment = Rs. 14,500

ii)The amount paid by him till 20th ins.talment

= Sum of all ins.talments from 1 to 20

We know that

Sum of first n terms = Sn= (n/2)[2a+(n-1)d]

We have , a = 5000, d = 500 and n = 20

=> S 20 = (20/2)[2(5000)+(20-1)(500)]

=> S 20 = (10)[10000+19×500)]

=> S 20 = 10(10000+9500)

=> S 20 = 10×19500

=> S 20 = 1,95,000

The amount paid by him till 20th inst.alment

= Rs. 1,95,000

iii)The amount paid by him till 30th ins.talment

= Sum of all ins.talments from 1 to 30

We know that

Sum of first n terms = Sn= (n/2)[2a+(n-1)d]

We have , a = 5000, d = 500 and n = 30

=> S 30 = (30/2)[2(5000)+(30-1)(500)]

=> S 30 = (15)[10000+29×500)]

=> S 30 = 15(10000+14500)

=> S 30 = 15×24500

=> S 30 =3,67,500

The amount paid by him till 30th ins.talment

= Rs. 3,67,500

iv) The amount still have to pay by him after 30th ins.talment =

Total Amount to be paid - The amount paid till 30th ins.talment

= 590000 - 367500

= Rs. 222500

Answer :-

i)The amount is paid by him in 20th ins.talment Rs. 14,500

ii)The amount paid by him till 20th ins.talment

= Rs. 1,95,000

iii)The amount paid by him till 30th ins.talment

= Rs. 3,67,500

iv)The amount still have to pay by him after 30th inst alment =Rs. 222500

Used formulae:-

  • nth term of an AP = an = a+(n-1)d.
  • Sum of first n terms = Sn= (n/2)[2a+(n-1)d]
  • a = First term
  • d = Common difference
  • n = Number of terms
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