Advertisements
Advertisements
Question
Solve the equation:
– 4 + (–1) + 2 + 5 + ... + x = 437
Advertisements
Solution
Given equation is,
– 4 + (–1) + 2 + 5 + … + x = 437
Here, –4 – 1 + 2 + 5 + … + x is in A.P.
Then, a = -4, d = -1 + 4 = 3, l = x
Given: Sn = 437
⇒ `"n"/2[2"a" + ("n" - 1)"d"]=437`
⇒ n[-8 + 3n - 3] = 874
⇒ 3n2 - 11n - 874 = 0
⇒ 3n2 - 57n + 46n - 874 = 0
⇒ 3n(n - 19) + 46(n -19) = 0
⇒ (n - 19) (3n + 46) = 0
⇒ n - 19 = 0, n = 19
⇒ 3n + 46 = 0
n = `-46/3` (Impossible)
Hence, n = 19
So, l = a + (n - 1)d
x = -4 + (19 - 1) 3
= -4 + 18 × 3
= -4 + 54
x = 50
APPEARS IN
RELATED QUESTIONS
Show that a1, a2,..., an... form an AP where an is defined as below:
an = 9 − 5n
Also, find the sum of the first 15 terms.
Find the sum of the following arithmetic progressions:
3, 9/2, 6, 15/2, ... to 25 terms
Find the sum of the first 51 terms of the A.P: whose second term is 2 and the fourth term is 8.
Find the sum of the first 22 terms of the A.P. : 8, 3, –2, ………
How many three-digit natural numbers are divisible by 9?
For what value of n, the nth terms of the arithmetic progressions 63, 65, 67, ... and 3, 10, 17, ... equal?
In an A.P., the sum of first ten terms is −150 and the sum of its next ten terms is −550. Find the A.P.
If the sum of first n terms of an A.P. is \[\frac{1}{2}\] (3n2 + 7n), then find its nth term. Hence write its 20th term.
Q.6
Rohan repays his total loan of ₹ 1,18,000 by paying every month starting with the first installment of ₹ 1,000. If he increases the installment by ₹ 100 every month, what amount will be paid by him in the 30th installment? What amount of loan has he paid after 30th installment?
