In an ap s10=136and s9=120then findt10
Answers
Step-by-step explanation:
Given:-
In an ap s10=136and s9=120
To find:-
find t10
Solution:-
In an AP, S10=136
S9=120
we know that Sum of n terms in an AP is denoted by Sn and defined by Sn=(n/2)[2a+(n-1)d
S10=(10/2)[2a+(10-1)d]
=>S10=5[2a+9d]
=>S10=10a+45d
According to the given problem,
S10=136
10a+45d=136----------(1)
S9=(9/2)[2a+(9-1)d]
=>S9=(9/2)[2a+8d]
=>S9=(9/2)×2(a+4d)
=>S9=9(a+4d)
=>S9=9a+36d
According to the given problem,
S9=120
9a+36d=120-----------(2)
On multiplying equation(1) with 9 then
90a+405d=1224------(3)
On multiplying equation (2) with 10 then
90a+360d=1200------(4)
On solving (3)&(4) then
Subtracting (4) from (3)
90a+405d=1224
90a+360d=1200
(-)
______________
0+45d=24
______________
we have,45d=24
=>d=24/45
=>d=8/15----------------(5)
On Substituting the value of d in the equation (1)
=>10a+45(8/15)=136
=>10a+3(8)=136
=>10a+24=136
=>10a=136-24
=>10a=112
=>a=112/10
=>a=56/5
we have,
First term (a)=56/5
Common difference (d)=8/15
Now, as we know that nth term in the AP
tn=a+(n-1)d
=>t10=(56/5)+(10-1)(8/15)
=>t10=(56/5)+9(8/15)
=>t10=(56/5)+72/15
=>t10=[(56×3)+72]/15
=>t10=(168+72)/15
=>t10=240/15
=>t10=16
Answer:-
10th term of the given AP =t10=16
Used formulae:-
"a" is the first term ,"d" is the common difference of an Arithmetic Progression then
- nth term =tn=a+(n-1)d
- Sum of n terms =Sn=(n/2)[2a+(n-1)d