Advertisements
Advertisements
Question
How many terms of the series 18 + 15 + 12 + ........ when added together will give 45?
Advertisements
Solution
Here, we find that
15 – 18 = 12 – 15 = –3
Thus, the given series is an A.P. with first term 18 and common difference –3.
Let the number of terms to be added be 'n'.
`S_n = n/2 [2a + (n - 1)d]`
`\implies 45 = n/2 [2(18) + (n - 1)(-3)]`
`\implies` 90 = n[36 – 3n + 3]
`\implies` 90 = n[39 – 3n]
`\implies` 90 = 3n[13 – n]
`\implies` 30 = 13n – n2
`\implies` n2 – 13n + 30 = 0
`\implies` n2 – 10n – 3n + 30 = 0
`\implies` n(n – 10) – 3(n – 10) = 0
`\implies` (n – 10)(n – 3) = 0
`\implies` n – 10 = 0 or n – 3 = 0
`\implies` n = 10 or n = 3
Thus, the required number of terms to be added is 3 or 10.
RELATED QUESTIONS
In an A.P., the sum of first n terms is `(3n^2)/2 + 13/2 n`. Find its 25th term.
Find the sum of 28 terms of an A.P. whose nth term is 8n – 5.
In a flower bed, there are 43 rose plants in the first row, 41 in second, 39 in the third, and so on. There are 11 rose plants in the last row. How many rows are there in the flower bed?
The sum of first three terms of an AP is 48. If the product of first and second terms exceeds 4 times the third term by 12. Find the AP.
Find where 0 (zero) is a term of the A.P. 40, 37, 34, 31, ..... .
Ramkali would need ₹1800 for admission fee and books etc., for her daughter to start going to school from next year. She saved ₹50 in the first month of this year and increased her monthly saving by ₹20. After a year, how much money will she save? Will she be able to fulfil her dream of sending her daughter to school?
Write the value of a30 − a10 for the A.P. 4, 9, 14, 19, ....
Write the sum of first n odd natural numbers.
Q.11
The sum of first five multiples of 3 is ______.
