A sum of ₹7680 is paid off in 12 instalments . Each instalment is ₹20 less than the preceding one . Find the amount of the first and last instalment.
Answers
Given,
Sum of money = Rs. 7680
Intallment numbers = 12
Each in s tallment each Rs. 20 less than the previous intallment
To find,
The amounts of first and last in s tallment.
Solution,
We can simply solve this mathematical problem by using the following mathematical process.
Here, we need to apply the AP series formula,
Sum of the terms = 7680
Number of terms = 12
Common difference = -20
First term = Let, x
According to the AP series formula,
12/2 [2x + (12-1)× -20] = 7680
6 [2x- 220] = 7680
2x-220 = 1280
2x = 1500
x = 750
First in s tallment= Rs. 750
Last in s tallment = 750+(12-1) × (-20) = 750-220 = 530
Hence, first and last in s tallments are Rs. 750 and Rs. 530 respectively
Answer:
let consider (Sn)-7680,(n) -12, (d) -20 .
Step-by-step explanation:
Solve By AP formula ,.
Sn= n/2 [2a+(n-1) -d]
7680 = 12/2 [2a (12- 1) -20]
7680= 6 [2a + 11(-20)]
7680 = 6 [2a - 220 ]
7680/6. =2a-220
1280 =2a -220
-2a = -220-1280
2a = -1500
-a = -1500/2
-a = -750
a = 750
putting a=750 in second formula
tn= a +( n -1 ) d
t12 =750+(12-1) -20
=750-220
=530