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Therefore, Senthil's house number is 35.Example 2.39 The sum of first n, 2n and 3n terms of an A.P. are SS, and S, respectively,Prove that S = 315,-5).om um of first man and 3n terms of an A.P. respectively then |
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Answer» Ram is 10 year more than twice the age of Rahim . Ram age is 56 year (take rahim age as z) Answer: 3(S_2-S_1)=S_33(S2−S1)=S3 Step-by-step explanation: It is given that the sum of first n ,2n and 3n terms of AP be s1, s2 and s3. To prove:S_3=3(S_2-S_1)S3=3(S2−S1) The sum of first n terms isdefined as S_1=S_n=\frac{n}{2}[2a+(n-1)d]S1=Sn=2n[2a+(n−1)d] S_2=S_{2n}=\frac{2n}{2}[2a+(2n-1)d]S2=S2n=22n[2a+(2n−1)d] S_3=S_{3n}=\frac{3n}{2}[2a+(3n-1)d]S3=S3n=23n[2a+(3n−1)d] Taking right hand side of the given equation. 3(S_2-S_1)=3(\frac{2n}{2}[2a+(2n-1)d]-\frac{n}{2}[2a+(n-1)d])3(S2−S1)=23n(2[2a+2dn−d]−[2a+dn−d]) 3(S_2-S_1)=\frac{3n}{2}(4a+4dn-2d-2a-dn+d)3(S2−S1)=23n(4a+4dn−2d−2a−dn+d) 3(S_2-S_1)=\frac{3n}{2}(2a+3dn-d)3(S2−S1)=23n(2a+3dn−d) 3(S_2-S_1)=\frac{3n}{2}(2a+(3n-1)d)3(S2−S1)=23n(2a+(3n−1)d) 3(S_2-S_1)=S_{3n}3(S2−S1)=S3n 3(S_2-S_1)=S_33(S2−S1)=S3 L.H.S=R.H.S. Hence prove. |
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