Question

# In an AP:(i) Given $$a=5, d=3$$, $$\displaystyle { a }_{ n }=50$$, find $$n$$ and $$\displaystyle { S }_{ n }$$.(ii) Given $$a=7$$, $$\displaystyle { a }_{ 13 }=35$$, find $$d$$ and $$\displaystyle { S }_{ 13 }$$.(iii) Given $$\displaystyle { a }_{ 12 }=37$$, $$d=3$$, find $$a$$ and $$\displaystyle { S }_{ 12 }$$.(iv) Given $$\displaystyle { a }_{ 3 }=15$$, $$\displaystyle { S }_{ 10 }=125$$, find $$d$$ and $$\displaystyle { a }_{ 10 }$$.(v) Given $$d=5$$, $$\displaystyle { S }_{ 9 }=75$$, find $$a$$ and $$\displaystyle { a }_{ 9 }$$.(vi) Given $$a=2, d=8$$, $$\displaystyle { S }_{ n }=90$$, find $$n$$ and $$\displaystyle { a }_{ n }$$.(vii) Given $$a=8$$, $$\displaystyle { a }_{ n }=62$$, $$\displaystyle { S }_{ n }=210$$, find $$n$$ and $$d$$.(viii) Given $$\displaystyle { a }_{ n }=4$$, $$d=2$$, $$\displaystyle { S }_{ n }=-14$$, 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 $$a$$.

Solution

## We know that in an APIf first term is $$a_1 = a$$, common difference is $$d$$ then the $$n^{th}$$ term is $$a_n = a+(n-1)d$$Sum of first $$n$$ terms of this  AP is $$S_n = \dfrac n2(a_1+a_n) = \dfrac n2 (2a+(n-1)d)$$Now,(i) $$a_1 = a = 5$$$$d = 3$$$$a_n = 50 = a + (n-1)d$$$$\Rightarrow 50 = 5 + (n-1)3$$$$\Rightarrow n = 16$$$$S_n = \dfrac n2(a_1 + a_n)$$$$\Rightarrow S_n = \dfrac{16}2(5+50)$$$$\Rightarrow S_n = 8\times 55 = 440$$(ii)$$a_1 = 7$$$$a_{13} = 35$$$$\Rightarrow n = 13$$$$a_{13} = a_1 + (n-1)d$$$$\Rightarrow 35 = 7 + (13-1)d$$$$\Rightarrow d = \dfrac73$$$$S_{13} = \dfrac {13}2(a_1 + a_{13})$$$$\Rightarrow S_{13} = \dfrac {13}2 (7+35)$$$$\Rightarrow S_{13} = 13\times 21 = 273$$(iii) $$a_{12} = 37$$$$d = 3$$$$\Rightarrow n = 12$$$$a_{12} = a_1 + (n-1)d = a + (12-1)3$$$$\Rightarrow 37 = a + 33$$$$\Rightarrow a = 4$$$$S_{12} = \dfrac n2 (a_1+a_{12})$$$$\Rightarrow s_{12} = \dfrac{12}2 (4 + 37) = 6\times 41 = 256$$(iv)$$a_3 = 15$$$$\Rightarrow a_3 = a + (3-1)d \Rightarrow 15 = a+ 2d$$    ...(1)$$S_{10} = 125$$$$\dfrac{10}2 (2a + 9d) = 125$$$$\Rightarrow 2a + 9d = 25$$    ....(2)From (1) and (2), we get$$d =-1, a = 17$$$$a_{10} = 17 + (9)(-1) = 8$$(v)$$d = 5$$$$S_9 = 75 = \dfrac 92 (2a + 8d) = \dfrac92 (2a + 40) = 9(a+20)$$$$\Rightarrow a = -\dfrac{35}3$$$$a_9 = a + 8d = -\dfrac{35}3 + 8\times 5 = 40 - \dfrac{35}3$$$$a_9 = \dfrac{85}3$$(vi)$$a=2$$$$d=8$$$$S_n = 90$$$$\Rightarrow \dfrac n2 (2a + (n-1)d) = 90$$$$\Rightarrow \dfrac n2(8n - 4) = 90$$$$\Rightarrow n(4n - 2) = 90$$$$\Rightarrow 2n^2 - n - 45=0$$$$\Rightarrow n = 5$$      $$(\because n>0, \therefore$$ we have neglected negative value of $$n = -\dfrac92)$$$$a_n = a_5 = a + (4)(d) = 2 + 4\times8 = 34$$(vii)$$a = 8$$$$a_n = 62$$$$S_n = 210 = \dfrac n2 (a + a_n)$$$$\Rightarrow 420 = n(8 + 62) = 70n$$$$\Rightarrow n = 6$$$$a_n = a + (n-1)d$$$$62 = 8 + (5)d$$$$\Rightarrow d = \dfrac{54}5$$(viii)$$a_n = 4$$$$d = 2$$$$a_n = a + (n-1)d$$$$4 = a + (n-1)2$$$$\Rightarrow = 6 = a+2n$$    ...(1)$$S_n = -14$$$$S_n = \dfrac n2 (a + a_n)$$$$-14 = \dfrac n2 (a + 4)$$    ....(2)From (1) and (2), we get$$n=7$$ or $$n = -2$$,  ($$\because n>0,$$ we have to neglect negative value)$$n = 7,$$ then $$a = -8$$(ix) $$a = 3$$$$n=8$$$$S = 192$$$$S = 192 = \dfrac n2 (2a + (n-1)d)$$$$\Rightarrow 48 = 6 + 7d$$$$\Rightarrow d = 6$$(x)$$l = a_n = 28$$$$S_n = 144$$$$n = 9$$$$S_n = \dfrac n2(a + a_n)$$$$144 = \dfrac 92 (a + 28)$$$$\Rightarrow a = 4$$MathematicsRS AgarwalStandard X

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