Advertisements
Advertisements
प्रश्न
If ₹ 3900 will have to be repaid in 12 monthly instalments such that each instalment being more than the preceding one by ₹ 10, then find the amount of the first and last instalment
Advertisements
उत्तर
The instalments are in A.P.
Amount repaid in 12 instalments (S12) = 3900
Number of instalments (n) = 12
Each instalment is more than the preceding one by ₹ 10.
∴ d = 10
Now, Sn = `"n"/2 [2"a" + ("n" - 1)"d"]`
∴ S12 = `12/2[2"a" + (12 - 1)(10)]`
∴ 3900 = 6[2a + 11(10)]
∴ 3900 = 6(2a + 110)
∴ `3900/6` = 2a + 110
∴ 650 = 2a + 110
∴ 2a = 540
∴ a = `540/2` = 270
tn = a + (n – 1)d
∴ t12 = 270 + (12 – 1)(10)
= 270 + 11(10)
= 270 + 110
= 380
∴ Amount of the first instalment is ₹ 270 and that of the last instalment is ₹ 380.
APPEARS IN
संबंधित प्रश्न
Find the sum of all numbers from 50 to 350 which are divisible by 6. Hence find the 15th term of that A.P.
Find the sum of first 30 terms of an A.P. whose second term is 2 and seventh term is 22
In an AP given a = 8, an = 62, Sn = 210, find n and d.
If the pth term of an A. P. is `1/q` and qth term is `1/p`, prove that the sum of first pq terms of the A. P. is `((pq+1)/2)`.
Find the sum of the following arithmetic progressions
`(x - y)^2,(x^2 + y^2), (x + y)^2,.... to n term`
In an A.P., if the 5th and 12th terms are 30 and 65 respectively, what is the sum of first 20 terms?
If the 10th term of an AP is 52 and 17th term is 20 more than its 13th term, find the AP
Find the 6th term form the end of the AP 17, 14, 11, ……, (-40).
Find the common difference of an AP whose first term is 5 and the sum of its first four terms is half the sum of the next four terms.
Find the sum of the first n natural numbers.
Write an A.P. whose first term is a and the common difference is d in the following.
a = 10, d = 5
The 9th term of an A.P. is equal to 6 times its second term. If its 5th term is 22, find the A.P.
The nth term of an A.P., the sum of whose n terms is Sn, is
Which term of the AP 3, 15, 27, 39, ...... will be 120 more than its 21st term?
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.
The sum of first 16 terms of the AP: 10, 6, 2,... is ______.
If the numbers n - 2, 4n - 1 and 5n + 2 are in AP, then the value of n is ______.
The sum of the first five terms of an AP and the sum of the first seven terms of the same AP is 167. If the sum of the first ten terms of this AP is 235, find the sum of its first twenty terms.
How many terms of the AP: –15, –13, –11,... are needed to make the sum –55? Explain the reason for double answer.
Find the sum of those integers between 1 and 500 which are multiples of 2 as well as of 5.
