Question

in an AP ; if S5+S7 =167 and S10 = 235 then find the AP ; where Sn denote the sum of its first n terms

Answer 1

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HERE IS YOUR ANSWER GOES LIKE THIS :


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YOUR QUESTION:

in an AP ;if S5+S7=167 and S10=235 then find the AP ; where Sn denote the sum of its first n terms

ANSWER:


let a and b be the first term common difference of the AP,respectively


therefore ,sum of n terms ,Sn =n/2(2a+(n-1)d)

we have 

S5+S7=167

⇒5/2(2a+4d)+7/2(2a+6d)=167

⇒5(a+2d)+7(a+3d)=167

⇒12a+31d=167 ----- (1)


also,

S10=235

⇒10/2(2a+9d)=23

⇒2a+9d=47--------(2)

solving (1) and (2) ,we get 

a=1 and d=5

Hence ,required AP is 1,6,11



HENCE THE ANSWER HAS BEEN EXPLAINED........


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❤️I HOPE THIS WILL HELP YOU ❤️


HAVE A GREAT DAY AHEAD !!!✌️✌️


^_^

Answer 2

HERE IS YOUR SOLUTION.......

LET FIRST TERM BE a AND DIFFERENCE BE d....

Sn = n/2 [ 2a + (n - 2) d ]
S5 + S7 = 167
5/2 [ 2a + (5 - 2) d ] + 7/2 [ 2a + ( 7 - 2) d ] = 167
5/2 [ 2a + 3d ] + 7/2 [ 2a + 5d ] = 167
10a +15d/2 + 14a + 35d/2 =. 167
24a + 50d = 167..........{1}

Sn = n/2 [ 2a + (n - 2) d ]
S10 = 235
10/2 [ 2a + ( 10 - 2 ) d ] = 235
10a + 40d = 235...........{2}

MULTIPLY EQN....2....BY 5 AND EQN...1....BY 4...

50a + 200d = 835......{3}
96a + 200d = 1336......{4}
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-46a = -501

FROM EQN..3 AND 4, WE GOT...
470 - 18d / 2 = 167 - 10d / 2
470 - 18d = 167 - 10d
-18d + 10d = 167 - 470
-8d = -303
d = 37.9.....

Substitute the value of d in EQN..1...

2a + 10d = 167
2a + 10 × 37.9 = 167
2a = 167 - 379
2a = -212
a = 106....

AP'S ARE ....106 , 143.9 , 181.8 , 219.7 , 257.6......n

I , HOPE THAT IT WILL HELP YOU A LOT.......

@ujjwalusri