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Misc 3 - Let sum of n, 2n, 3n terms of AP be S1, S2, S3 - Miscellaneous

Misc 3 - Chapter 9 Class 11 Sequences and Series - Part 2
Misc 3 - Chapter 9 Class 11 Sequences and Series - Part 3


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Misc 3 Let the sum of n, 2n, 3n terms of an A.P. be S1, S2 and S3, respectively . Show that S3 = 3 (S2– S1) We know that Sum of n terms = n/2 [2a + (n – 1)d] i.e. Sn = n/2 [2a + (n – 1)d] where a is the first term & d is the common difference of the A.P. It is given that Sum of n terms = S1 i.e. n/2 [2a+(n – 1)d] = S1 Also, Sum of 2n terms = S2 i.e. 2n/2 [2a + (2n – 1)d] = S2 And Sum of 3n terms = S3 i.e. 3n/2 [2a + (3n – 1)d] = S3 We want to show S3 = 3(S2 – S1) Taking R.H.S 3(S2 – S1) = 3[2n/2 " [2a + (2n" −"1)d]" −n/2 " [2a + (n" −"1)d]" ] = 3 (n/2)["2[2a + (2n" −"1)d]" −"[2a + (n" −"1)d]" ] = 3n/2 ["2(2a) + 2(2n" −"1)d" −"2a" −"(n" −"1)d" ] = 3n/2 ["4a + (4n" −"2)d" −"2a" −"(n" −"1)d" ] = 3n/2 ["4a + 4nd" −"2d" −"2a" −"nd + d" ] = 3n/2 ["4a" −"2a + 4nd" −"nd" −"2d + d" ] = 3n/2 ["2a + 3nd" −"d" ] = 3n/2 ["2a + (3n – 1)d" ] From (3), S3 = 3𝑛/2 [2a + (3n – 1)d = S3 = L.H.S Hence, R.H.S = L.H.S Hence proved

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.