Math, asked by yasiraffan, 11 months ago

In an AP, it is given that S5+ S7 = 167 and S10 = 235, then find the AP, where Sn denotes the sum of its first n terms.​

Answers

Answered by kartik2507
6

Answer:

1, 6, 11, 16..........

Step-by-step explanation:

S5 + S7 = 167

Sn = n/2 (2a + (n-1)d)

 \frac{5}{2} (2a + (5 - 1)d) +   \frac{7}{2}(2a + (7 - 1)d) = 167 \\  \frac{5}{2}  (2a + 4d) +  \frac{7}{2}  (2a + 6d) = 167 \\  \frac{5}{2}   \times 2(a + 2d) +  \frac{7}{2}  \times 2(a + 3d) = 167 \\ 5a + 10d + 7a + 21d = 167 \\ 12a + 31d = 167 \:  \:  \:  \: equ(1) \\ \\  s10 =  \frac{10}{2} (2a + 9d) = 235 \\   5(2a + 9d) = 235  \\ 2a + 9d =  \frac{235}{5} \\ 2a + 9d = 47  \:  \:  \:  \: equ(2) \\  \\ multiply \: equ \: (2) \: with \: 6 \\ 12a + 54d = 282 \:  \:  \:  \: equ(3) \\ equ \: (3) - (1) \\ 23d = 115 \\ d =  \frac{115}{23}  \\ d = 5

substitute d = 5 in equ 1

substitute d = 5 in equ 2

2a + 9d = 47

2a + 9(5)= 47

2a + 45 = 47

2a = 47 - 45

2a = 2

a = 2/2

a = 1

the AP is

1, 6, 11, 16 ..........

hope you get your answer

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