Advertisements
Advertisements
प्रश्न
A merchant borrows ₹ 1000 and agrees to repay its interest ₹ 140 with principal in 12 monthly instalments. Each instalment being less than the preceding one by ₹ 10. Find the amount of the first instalment.
Advertisements
उत्तर
The installments are in A.P.
Amount repaid in 12 instalments (S12)
= Amount borrowed + Total interest
= 1000 + 140
∴ S12 = 1140
Number of instalments (n) = 12
Each instalment is less than the preceding one by ₹ 10.
∴ d = –10
Now, `S_n = n/2 [2a + (n - 1)d]`
∴ `S_12 = 12/2 [2a + (12 - 1)(-10)]`
∴ 1140 = 6[2a + 11(– 10)]
∴ 1140 = 6(2a – 110)
∴ `1140/6` = 2a – 110
∴ 190 = 2a – 110
∴ 2a = 300
∴ a = `300/2`
∴ a = 150
∴ The amount of first instalment is ₹ 150.
संबंधित प्रश्न
If the ratio of the sum of first n terms of two A.P’s is (7n +1): (4n + 27), find the ratio of their mth terms.
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Find the 12th term from the end of the following arithmetic progressions:
3, 5, 7, 9, ... 201
Find the sum of all natural numbers between 1 and 100, which are divisible by 3.
Write an A.P. whose first term is a and common difference is d in the following.
a = –19, d = –4
In an A.P. 19th term is 52 and 38th term is 128, find sum of first 56 terms.
If the seventh term of an A.P. is \[\frac{1}{9}\] and its ninth term is \[\frac{1}{7}\], find its (63)rd term.
The sum of 5th and 9th terms of an A.P. is 30. If its 25th term is three times its 8th term, find the A.P.
The sum of first n terms of an A.P. is 3n2 + 4n. Find the 25th term of this A.P.
If the sum of n terms of an A.P. is 2n2 + 5n, then its nth term is
If the first term of an A.P. is 2 and common difference is 4, then the sum of its 40 terms is
If the sums of n terms of two arithmetic progressions are in the ratio \[\frac{3n + 5}{5n - 7}\] , then their nth terms are in the ratio
Q.14
The 11th term and the 21st term of an A.P are 16 and 29 respectively, then find the first term, common difference and the 34th term.
Show that a1, a2, a3, … form an A.P. where an is defined as an = 3 + 4n. Also find the sum of first 15 terms.
Find the sum of three-digit natural numbers, which are divisible by 4.
Find the sum:
`(a - b)/(a + b) + (3a - 2b)/(a + b) + (5a - 3b)/(a + b) +` ... to 11 terms
If the last term of an A.P. of 30 terms is 119 and the 8th term from the end (towards the first term) is 91, then find the common difference of the A.P. Hence, find the sum of all the terms of the A.P.
The sum of 40 terms of the A.P. 7 + 10 + 13 + 16 + .......... is ______.
