Advertisements
Advertisements
Question
If the third term of an A.P. is 1 and 6th term is – 11, find the sum of its first 32 terms.
Advertisements
Solution
T3 = 1, T6 = –11, n = 32
a + 2d = 1 ...(i)
a + 5d = –11 ...(ii)
– – +
Subtracting (i) and (ii),
–3d = 12
⇒ d = `(12)/(-3)` = –4
Substitute the value of d in eq. (i)
a + 2(–4) = 1
⇒ a – 8 = 1
a = 1 + 8 = 9
∴ a = 9, d = –4
S32 = `n/(2)[2a + (n - 1)d]`
= `(32)/(2)[2 xx 9 + (32 - 1) xx (–4)]`
= 16[18 + 31 x (–4)]
= 16[18 – 124 ]
= 16 x (–106)
= –1696.
APPEARS IN
RELATED QUESTIONS
In an AP Given a12 = 37, d = 3, find a and S12.
The first and last terms of an AP are a and l respectively. Show that the sum of the nth term from the beginning and the nth term form the end is ( a + l ).
If the sum of first m terms of an AP is ( 2m2 + 3m) then what is its second term?
Find the sum (−5) + (−8)+ (−11) + ... + (−230) .
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?
The number of terms of the A.P. 3, 7, 11, 15, ... to be taken so that the sum is 406 is
Suppose three parts of 207 are (a − d), a , (a + d) such that , (a + d) >a > (a − d).
Mrs. Gupta repays her total loan of Rs. 1,18,000 by paying installments every month. If the installments for the first month is Rs. 1,000 and it increases by Rs. 100 every month, What amount will she pays as the 30th installments of loan? What amount of loan she still has to pay after the 30th installment?
Q.16
Solve the equation
– 4 + (–1) + 2 + ... + x = 437
