CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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 AP

If 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)
$$l = a_n = 28$$

$$S_n = 144$$

$$n = 9$$

$$S_n = \dfrac n2(a + a_n)$$

$$144 = \dfrac 92 (a + 28)$$

$$\Rightarrow a = 4$$

Mathematics
RS Agarwal
Standard X

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
Same exercise questions
View More


similar_icon
People also searched for
View More



footer-image