a man repaysa loan for 3250 by paying rupees 20 in the first month and then increasing the payments Rupees 15 every month how many long will be
Answers
Answered by
1
This is in Arithmetic Progression
Here a=20, d = 15 ,
3250 = n/2[40+(n-1)15]
∵ sum of series S_{n} [/tex] = n/2[2a+(n-1)d]
3250 = n/2[25+15n]
6500 = 25n+15n²
⇒15n²+25n-6500= 0
⇒ (6n+130)(n-20) = 0
⇒x = -130/6, 20
∵ the value cant be negative, so months n = 20
Here a=20, d = 15 ,
3250 = n/2[40+(n-1)15]
∵ sum of series S_{n} [/tex] = n/2[2a+(n-1)d]
3250 = n/2[25+15n]
6500 = 25n+15n²
⇒15n²+25n-6500= 0
⇒ (6n+130)(n-20) = 0
⇒x = -130/6, 20
∵ the value cant be negative, so months n = 20
Similar questions