Check sibling questions

Ex 9.2, 17 - A man starts repaying a loan as first instalment - Arithmetic Progression (AP): Statement

Ex 9.2, 17 - Chapter 9 Class 11 Sequences and Series - Part 2
Ex 9.2, 17 - Chapter 9 Class 11 Sequences and Series - Part 3

This video is only available for Teachoo black users

Get Real time Doubt solving from 8pm to 12 am!


Transcript

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

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.