Sum of first k terms of an AP is given by 7k – 2k²
. If the pth term is –35, then find the value
of p. Which term of the AP is –99
Answers
Step-by-step explanation:
Given:-
Sum of first k terms of an AP is given by
7k – 2k^2 and Pth term is -35
To find:-
1)find the value of P
2)Which term of the AP is –99
Solution:-
Given that
Sum of first k terms of an AP = 7k - 2k^2
Sk = 7k - 2k^2
If k = 1 then
=>S1 = 7(1)-2(1)^2
=>S1 = 7-2 = 5
First term of the AP = 5
If k = 2 then
=>S2 = 7(2)-2(2)^2
=>S2 = 14-2(4)
=>S2 = 14-8
=>S2 = 6
Sum of two terms = 6
=>First term + Second term = 6
=>5+Second term = 6
=>Second term = 6-5
Second term = 1
Common difference = second term - First term
=>d = 1-5
=>d = -4
Now
As we know
nth term of an AP is an
an = a+(n-1) d
Now Pth term = -35
we have a = 5 ,d = -4, ap = -35
=>ap = a+(p-1)d
=>-35 = 5+(p-1)(-4)
=>-35 = 5 -4p +4
=>-35 = 9 -4p
=>-35-9 = -4p
=>-4p = -44
=>p = -44/-4
=>p = 11
The value of p = 11
Let -99 is the nth term of the AP
an = -99
=>a+(n-1)d = -99
=>5+(n-1)(-4) = -99
=>5-4n+4 = -99
=>9-4n = -99
=>-4n = -99-9
=>-4n = -108
=>4n = 108
=>n = 108/4
=>n = 27
27th term of the AP is -99
Answer:-
The value of p for the given problem is 11
-99 is the 27th term of the given AP
Used formulae:-
- General term of the AP is an = a+(n-1) d
Where, a is the first term
n is the number of terms
d is the common difference of the AP