Math, asked by benetta, 1 year ago

find first four terms of sequence of which sum to n terms is 1/2n (7n-1)

Answers

Answered by abhi178
43
Given, \bold{S_n=\frac{1}{2}n(7n-1)}

We know, one important concept,
\bold{T_n=S_n-S_{n-1}} use it here,
\bold{S_n=\frac{1}{2}n(7n-1)}\\\\\bold{S_{n-1}=\frac{1}{2}(n-1)(7n-8)}
So, \bold{T_n=\frac{1}{2}n(7n-1)-\frac{1}{2}(n-1)(7n-8)}\\\\=\bold{\frac{1}{2}[7n^2-n-7n^2+15n-8]}\\\\=\bold{\frac{1}{2}(14n-8)}\\\\=\bold{7n-4}

Now, Tn = 7n - 4
First term , n = 1 ⇒T₁ = 7 × 1 - 4 = 3
second term , n = 2 ⇒, T₂ = 7 × 2 - 4 = 10
3rd term , n = 3 ⇒T₃ = 7 × 3 - 4 = 17
4th term , n = 4 ⇒T₄ = 7 × 4 - 4 = 24

ÄrnoVictorDoriän: can you explain Sn-1 = 1/2(n-1)(7n-8)
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