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Answers
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Let the first term is a and the common difference is d
By using S
n
=
2
n
[2a+(n−1)d] we have,
S
5
=
2
5
[2a+(5−1)d]=
2
5
[2a+4d]
S
7
=
2
7
[2a+(7−1)d]=
2
7
[2a+6d]
Given: S
7
+S
5
=167
∴
2
5
[2a+4d]+
2
7
[2a+6d]=167
⇒10a+20d+14a+42d=334
⇒24a+62d=334 ...(1)
S
10
=
2
10
[2a+(10−1)d]=5(2a+9d)
Given: S
10
=235
So 5(2a+9d)=235
⇒2a+9d=47 ...(2)
Multiply equation (2) by 12, we get
24a+108d=564....(3)
Subtracting equation (3) from (1), we get
−46d=−230
∴d=5
Substing the value of d=5 in equation (1) we get
2a+9(5)=47 or 2a=2
∴a=1
Then A.P is 1,6,11,16,21,⋯
Step-by-step explanation:
S₅+S₇=167 and S₁₀=235
Now, Sₙ=n/2{2a+(n-1)d}
∴S₅+S₇=167
⇒5/2{2a+4d}+7/2{2a+6d}
⇒5a+10d+7a+21d=167
⇒12a+31d=167 ..... (1)
Also , S₁₀=235
∴10/2{2a+9d}=235
⇒10a+45d=235
⇒2a+9d=47 .....(2)
Multiplying equation (2) by 6, we get
12a+54d=282 .....(3)
subracting (1) from (3) we get
12a+54d=282
(-) 12a+31d=167
- - -
-----------------------
23d=115
--------------------------
∴d=5
substituting value of d in (2), we have
2(a)+9(5)=47
⇒2a+45=47
⇒2a=2
⇒a=1
Thus AP is 1 , 6 , 11 , 16.........