![Ex 9.2, 17 - Chapter 9 Class 11 Sequences and Series - Part 2](https://d1avenlh0i1xmr.cloudfront.net/74118301-5977-41f9-aeb1-d66a785694ccslide6.jpg)
![Ex 9.2, 17 - Chapter 9 Class 11 Sequences and Series - Part 3](https://d1avenlh0i1xmr.cloudfront.net/6a4eb5cc-285e-4194-8c7d-c0e021058888slide7.jpg)
Arithmetic Progression
Arithmetic Progression
Last updated at April 16, 2024 by Teachoo
Ex9.2,17 A man starts repaying a loan as first instalment of Rs.100. If he increases the instalment by Rs 5 every month, what amount he will pay in the 30th instalment? First instalment = 100. Every month instalment increases by Rs 5 Second instalment = 100 + 5 = 105 Third instalment = 105 + 5 = 110 So, the instalments are 100, 105, 110, The instalments is in A.P as difference between consecutive terms is constant. 100, 105, 110, Here, first term = a = 100 & common difference = 105 100 = 5 We need to find 30th instalment To find it, we use the formula an = a + (n 1)d where an = nth term , n = number of terms, a = first term , d = common difference Here, an = 30th instalment , n = 30 , a = 100, d = 5 a30 = 100 + (30 1)5 = 100 + (29) 5 = 100 + 145 = 245 Hence, the 30th instalment is Rs 245