Let S10 denote the sum of the first 10 terms of an A.P. If S20=3S10 then S30:S10 is equal to....

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Let S denote sum of first n terms of an arithmetic progression.If S20 790 and S10 145 then S15-S5 is Let S denote sum of first n terms of an arithmetic progression.If S20 790 and S10 145 then S15-S5 is
If Sn denotes the sum of first n terms of an AP, prove that S30=3[S20-S10] (CBSE 2015) 30/1/3 If Sn denotes the sum of first n terms of an AP, prove that S30=3[S20-S10] (CBSE 2015) 30/1/3
If Sn denotes the sum of the first n terms of an A.P., prove that S30 = 3(S20-S10). If Sn denotes the sum of the first n terms of an A.P., prove that S30 = 3(S20-S10).
If Sn​ denotes the sum of the first n terms of an A.P., prove that S30​=3(S20​−S10​) If Sn​ denotes the sum of the first n terms of an A.P., prove that S30​=3(S20​−S10​)
Let Sn denote the sum of first n terms of an AP. If S10=530 and S5=140 then S20-S6 is equal to Let Sn denote the sum of first n terms of an AP. If S10=530 and S5=140 then S20-S6 is equal to
Let Sn denote the sum of first n-terms of an AP. If S20 = 790, S10 = 145, then S15 – S5 is equal to Let Sn denote the sum of first n-terms of an AP. If S20 = 790, S10 = 145, then S15 – S5 is equal to
When sum of first 10 terms S_10 = 100, first 20 terms S_20 = 400 find the sum of the first 30 terms When sum of first 10 terms S_10 = 100, first 20 terms S_20 = 400 find the sum of the first 30 terms
In an arithmetic progression, if S40 = 1030 and S12 = 57, then S30 – S10 is equal to: In an arithmetic progression, if S40 = 1030 and S12 = 57, then S30 – S10 is equal to: