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Answer :

From the given statements, we can write,

\(a_3 + a_7 = 6\) …………………………….(i)

And

\(a_3 ×a_7 = 8\) ……………………………..(ii)

By the \(n^{th}\) term formula,

\(a_n = a+(n - 1)d\)

Third term, \(a_3 = a+(3 -1)d\)

\(a_3 = a + 2d\)………………………………(iii)

And Seventh term, \(a_7= a+(7-1)d\)

\(a_7 = a + 6d \)………………………………..(iv)

From equation (iii) and (iv), putting in equation(i), we get,

\(\Rightarrow \) a+2d +a+6d = 6

\(\Rightarrow \) 2a+8d = 6

\(\Rightarrow \) a+4d=3

\(\Rightarrow \) a = 3–4d …………………………………(v)

Again putting the eq.(iii) and (iv), in eq. (ii), we get,

(a+2d)×(a+6d) = 8

Putting the value of a from equation (v), we get,

\(\Rightarrow (3–4d +2d)×(3–4d+6d) = 8\)

\(\Rightarrow (3 –2d)×(3+2d) = 8\)

\(\Rightarrow 3^2 – 2d^2\) = 8

\(\Rightarrow 9 – 4d^2\) = 8

\(\Rightarrow 4d^2 = 1\)

\(\Rightarrow d = {{1} \over {2}} or -{{1} \over {2}}\)

Now, by putting both the values of d, we get,

\(\Rightarrow a = 3 – 4d = 3 – 4({{1} \over {2}}) = 3 – 2 = 1\), when \(d = {{1} \over {2}}\)

\(a = 3 – 4d = 3 – 4(-{{1} \over {2}}) = 3+2 = 5\), when \(d = -{{1} \over {2}}\)

We know, the sum of nth term of AP is;

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

So, when a = 1 and d=\({{1} \over {2}}\)

Then, the sum of first 16 terms are;

\(S_{16} = {{16} \over {2}} [2 +(16-1){{1} \over {2}}] \)

\(= 8(2+{{15} \over {2}}) = 76\)

And when a = 5 and d= -\({{1} \over {2}}\)

Then, the sum of first 16 terms are;

\(S_{16} = {{16} \over {2}} [2.5+(16-1)(-{{1} \over {2}})] \)

\( = 8({{5} \over {2}})=20\)

- Which term of the AP : 121, 117, 113, . . ., is its first negative term? [Hint : Find n for an < 0]
- A ladder has rungs 25 cm apart.. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are \({2} {\dfrac{1}{2}}\) apart, what is the length of the wood required for the rungs? [Hint : Number of rungs = -250/25 ].
- The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x. [Hint :Sx – 1 = S49 – Sx]
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