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# In an AP (i)Given a = 5, d = 3, an = 50, find n and $$S_n$$. (ii) Given a = 7, $$a_{13}$$ = 35, find d and $$S_{13}$$. (iii) Given $$a_{12}$$ = 37, d = 3, find a and $$S_{12}$$. (iv) Given $$a_3$$ = 15, $$S_{10}$$ = 125, find d and $$a_{10}$$. (v) Given d = 5, $$S_9$$= 75, find a and $$a_9$$. (vi) Given a = 2, d = 8, $$S_n$$ = 90, find n and an. (vii) Given a = 8, an = 62, $$S_n$$ = 210, find n and d. (viii) Given $$a_n$$ = 4, d = 2, $$S_n$$ = - 14, find n and a. (ix) Given a = 3, n = 8, S = 192, find d. (x)Given l = 28, S = 144 and there are total 9 terms. Find a.

(i) Given that, a = 5, d = 3, $$a_n$$ = 50

As we know, from the formula of the nth term in an AP,

$$a_n = a + (n – 1)d$$

Therefore, putting the given values, we get,

$$\Rightarrow$$ 50 = 5+(n -1)×3
$$\Rightarrow$$ 3(n -1) = 45
$$\Rightarrow$$ n -1 = 15
$$\Rightarrow$$ n = 16

Now, sum of n terms,

$$\Rightarrow S_n = (a + a_n)$$
$$\Rightarrow S_n = {{16} \over{2}} (5 + 50) = 440$$

(ii) Given that, a = 7, $$a_{13}$$ = 35

As we know, from the formula of the nth term in an AP
$$a_n = a + (n – 1)d$$

Therefore, putting the given values, we get,
$$\Rightarrow$$ 35 = 7+(13-1)d
$$\Rightarrow$$ 12d = 28
$$\Rightarrow$$ d = 28/12 = 2.33

Now, sum of n terms,

$$\Rightarrow S_n = (a + a_n)$$
$$\Rightarrow S_n = {{13} \over {2}} (7 + 35) = 273$$

(iii)Given that, $$a_{12}$$ = 37, d = 3

As we know, from the formula of the nth term in an AP,
$$a_n = a + (n – 1)d$$

Therefore, putting the given values, we get,
$$\Rightarrow a_{12}$$ = a + (12 - 1)3
$$\Rightarrow$$ 37 = a + 33
$$\Rightarrow$$ a = 4

Now, sum of nth term,

$$\Rightarrow S_n = (a + a_n)$$
$$\Rightarrow S_n = {{12} \over {2}} (4 + 37)$$
= 246

(iv) Given that, $$a_3$$ = 15, $$S_{10}$$ = 12

As we know, from the formula of the nth term in an AP,
$$a_n = a + (n – 1)d$$
Therefore, putting the given values, we get,

$$\Rightarrow a_3$$ = a+(3 - 1)d
$$\Rightarrow$$ 15 = a + 2d ………………………….. (i)

Sum of the nth term,

$$\Rightarrow S_n = {{n} \over {2}} [2a+(n-1)d]$$
$$\Rightarrow S_{10} = {{10} \over {2}} [2a+(10-1)d]$$
$$\Rightarrow$$ 125 = 5(2a+9d)
$$\Rightarrow$$ 25 = 2a+9d ……………………….. (ii)

On multiplying equation (i) by 2, we will get;
30 = 2a+4d ………………………………. (iii)

By subtracting equation (iii) from (ii), we get,
-5 = 5d
d = -1

From equation (i)
15 = a + 2(-1)
15 = a - 2

a = 17 = First term
$$\Rightarrow a_{10}$$ = a + (10-1)d
$$\Rightarrow a_{10}$$ = 17 + (9)(-1)
$$\Rightarrow a_{10}$$ = 17 -9 = 8

(v) Given that, d = 5, $$S_9$$ = 75

As, sum of n terms in AP is,
$$S_n = {{n} \over {2}} [2a+(n-1)d]$$

Therefore, the sum of first nine terms are;

$$\Rightarrow S_9 = {{9} \over {2}} [2a+(9 -1)5]$$
$$\Rightarrow$$ =25 = 3(a+20)
$$\Rightarrow$$ 25 = 3a+60
$$\Rightarrow$$ 3a = 25-60
$$\Rightarrow$$ a = $${{-35} \over {3}}$$

As we know, the nth term can be written as;
$$a_n = a + (n – 1)d$$
=$$a_9 = a + (9 - 1)(5)$$
= $${{-35} \over {3}}$$ +8(5)
= $${{-35} \over {3}}$$+40
= ($${{-35 + 120} \over {3}}$$)
= $${{85} \over {3}}$$

(vi) Given that, a = 2, d = 8, $$S_n$$ = 90

As, sum of n terms in an AP is,

$$\Rightarrow S_n = {{n} \over {2}} [2a+(n-1)d]$$
$$\Rightarrow 90 = {{n} \over {2}} [2a +(n -1)d]$$
$$\Rightarrow 180 = n(4 +8n - 8)$$
$$\Rightarrow 180 = n(8n-4) = 8n^2-4n$$
$$\Rightarrow 8n^2 - 4n –180 = 0$$
$$\Rightarrow 2n^2 - n - 45 = 0$$
$$\Rightarrow 2n^2 -10n + 9n - 45 = 0$$
$$\Rightarrow 2n(n - 5) + 9(n - 5) = 0$$
$$\Rightarrow(2n - 9)(2n + 9) = 0$$

So, n = 5 (as it is positive integer)
$$a_5$$ = 8 + 5×4 = 34

(vii) Given that, a = 8, $$a_n$$ = 62, $$S_n$$ = 210
As, sum of n terms in an AP is,

$$\Rightarrow S_n = {{n} \over {2}} [2a+(n-1)d]$$
$$\Rightarrow 210 = {{n} \over {2}} (8 +62)$$
$$\Rightarrow$$ 35n = 210
$$\Rightarrow$$ n = $$\frac{210}{35}$$ = 6

Now, 62 = 8 + 5d
$$\Rightarrow$$ 5d = 62 - 8 = 54
$$\Rightarrow$$ d = $${{54} \over {5}}$$ = 10.8

(viii) Given that, nth term, an = 4,
common difference, d = 2, sum of n terms, $$S_n$$ = -14.

As we know, from the formula of the nth term in an AP,
$$a_n = a + (n – 1)d$$

Therefore, putting the given values, we get,

$$\Rightarrow$$ 4 = a+(n -1)2
$$\Rightarrow$$ 4 = a+2n-2
$$\Rightarrow$$ a+2n = 6
$$\Rightarrow$$ a = 6 - 2n ……………………. (i)

As we know, the sum of n terms is;

$$\Rightarrow S_n = \frac{n}{2} (a + a_n)$$
$$\Rightarrow -14 = \frac{n}{2}(a+4)$$
$$\Rightarrow-28 = n(a + 4)$$
$$\Rightarrow-28 = n(6 -2n + 4)$${From equation (i)}
$$\Rightarrow-28 = n(-2n +10)$$
$$\Rightarrow-28 = - 2n^2 + 10n$$
$$\Rightarrow 2n^2 -10n - 28 = 0$$
$$\Rightarrow n^2 -5n -14 = 0$$
$$\Rightarrow n^2 -7n + 2n -14 = 0$$
$$\Rightarrow n(n - 7) + 2(n - 7) = 0$$
$$\Rightarrow (n -7)(n +2) = 0$$

Either n - 7 = 0 or n + 2 = 0
n = 7 or n = -2

However, n can neither be negative nor fractional.
Therefore, n = 7

From equation (i), we get
$$\Rightarrow$$ a = 6 - 2n
$$\Rightarrow$$ a = 6 - 2(7)
$$\Rightarrow$$ = 6 - 14

Thus, a = -8

(ix) Given that, first term, a = 3,
Number of terms, n = 8
And sum of n terms, S = 192
As we know,

$$\Rightarrow S_n = {{n} \over {2}} [2a+(n-1)d]$$
$$\Rightarrow 192 = {{8} \over {2}} [2×3+(8 -1)d]$$
$$\Rightarrow$$ 192 = 4[6 + 7d]
$$\Rightarrow$$ 48 = 6 + 7d
$$\Rightarrow$$ 42 = 7d
$$\Rightarrow$$ d = 6

(x) Given that, l = 28,S = 144 and there are total of 9 terms.

Sum of n terms formula,

$$\Rightarrow S_n = {{n} \over {2}} (a + l)$$
$$\Rightarrow 144 = {{9} \over {2}}(a+28)$$
$$\Rightarrow$$ (16)×(2) = a + 28
$$\Rightarrow$$ 32 = a + 28
$$\Rightarrow$$ a = 4