Question

if the sum of 7th term of an arithemetic progression is 182. if its 4th and 7th term are in ratio 1:3 find the ap​



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Answer 1

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Answer 2

Given, seventh Term of A.P

S7 = a+ (n- 1)d =182

➡️a+(7-1) d. =182

➡️a+6d = 182 ....(1)

Also,

S4/S7= 1/3

➡️ a+3d / a+6d=1/3

➡️ a+3d /182 =1/3

➡️ a+3d =182/3 .....(2)

By subtracting eq^n 2 by 1, we get,

eq^n (1) ❌1➡️ a+6d =182

(2) ❌1➡️ a+3d =182/3

(-). (-). (-)

6d- 3d =182 -182/3

➡️. 3d = 546- 182 /3.

➡️ 9d = 364

➡️. d= 40.44

Substituting the value of d in eq^n 1, we may also substitute it in 2 the answer would be same, we get,

➡️ a+6 ❌ 364 /9 =182

➡️ a +242.66 =182.

➡️ a= -60.66

Hence, A .P of the following series is ,a ,a+d, a+2d

i.e. -60.66, -60.66 + 40.44, -60.66+ 2❌40.44

➡️ -61 , -20 , 20 ,..... approx.