Math, asked by ubshinde1977, 2 months ago

In an ap s10=136and s9=120then findt10​

Answers

Answered by tennetiraj86
3

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
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