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(i) \(a_n = 3+4_n\)

(ii) \(a_n = 9 - 5_n\)

Also find the sum of the first 15 terms in each case.

Answer :

(i) \(\because a_n = 3+4_n\)

\( a_1 = 3+4(1) = 7\)

\( a_2 = 3+4(2) = 3+8 = 11\)

\( a_3 = 3+4(3) = 3+12 = 15\)

\( a_4 = 3+4(4) = 3+16 = 19\)

We can see here, the common difference between the terms are;

\( a_2 - a_1 = 11-7 = 4\)

\( a_3 - a_2 = 15-11 = 4\)

\( a_4 - a_3 = 19-15 = 4\)

Hence, \(a_k + 1 - a_k\) is the same value every time. Therefore, this is an AP with common difference as 4 and first term as 7.

Now, we know, the sum of \(n^{th}\) term is;

\(S_n = {{n} \over {2}}[2a+(n -1)d]\)

\(\therefore S_{15} = {{15} \over {2}}[2(7)+(15-1)×4]\)

= \({{15} \over {2}}[(14)+56]\)

= \({{15} \over {2}}(70)\)

= 15×35

= 525

(ii) \(a_n = 9 - 5n\)

\(a_1 = 9-5×1 = 9-5 = 4\)

\(a_2 = 9-5×2 = 9-10 = -1\)

\(a_3 = 9-5×3 = 9-15 = -6\)

\(a_4 = 9-5×4 = 9-20 = -11\)

We can see here, the common difference between the terms are;

\(a_2 - a_1 = -1-4 = -5\)

\(a_3 - a_2 = -6-(-1) = -5\)

\(a_4 - a_3 = -11-(-6) = -5\)

Hence, \(a_k + 1 - a_k\) is same every time. Therefore, this is an A.P. with common difference as -5 and first term as 4.

Now, we know, the sum of nth term is;

\(S_n = {{n} \over {2}} [2a +(n-1)d]\)

\(S_{15} = {{15} \over {2}}[2(4) +(15 -1)(-5)]\)

= \({{15} \over {2}}\)[8 + 14(-5)]

= \({{15} \over {2}}\)(8 - 70)

= \({{15} \over {2}}\)(-62)

= 15(-31)

= -465

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