Arithmetic Progressions Test - 40

Let  $$a$$ and $$d$$ be the first term and the common difference of the given AP respectively. Then,
$$a_6=19\Rightarrow a+5d=19$$                    $$...(1)$$
$$a_{17}=41\Rightarrow a+16d=41$$               $$ ...(2)$$
On subtracting $$(2)$$ from $$(1),$$ we get
$$11d=22\Rightarrow d=2$$
Putting $$d=2$$ in $$(1),$$ we get
$$a=19-5\times2=9$$
Now, $$a_{40}=a+39d=9+39\times2=87$$              

Let  $$a$$ and $$d$$ be the first term and the common difference of the given AP respectively. Then,
$$a_6=12\Rightarrow a+5d=12$$                    ...(1)
$$a_8=22\Rightarrow a+7d=22$$                ...(2)
On subtracting (2) from (1), we get
$$2d=10\Rightarrow d=5$$
Putting $$d=5$$ in (1), we get
$$a=12-5\times5=-13$$
Now, $$a_2=a+d=-13+5=-8$$ and
$$a_n=a+(n-1)d$$

$${\textbf{Step 1: Write multiple of 9 as A}}{\text{.P}}{\textbf{. having common difference as 9.}}$$

               $${\text{Multiple of 9 between 300 and 700 are as follows:}}$$

               $$306,315,324, \ldots ,693$$

               $${\text{So, the formed A}}{\text{.P}}{\text{. is}}$$ $$306,315,324, \ldots ,693$$

               $${\text{Where,}}$$ $$a = 306:$$ $${\text{First term of A}}{\text{.P}}{\text{.}}$$

               $$d  = 9:$$ $${\text{common difference of A}}{\text{.P}}{\text{.}}$$

               $$l = 639:$$ $${\text{Last term of A}}{\text{.P}}{\text{.}}$$

               $${\text{Now we know that, formula for }}{{\text{n}}^{{\text{th}}}}{\text{ term of an A}}{\text{.P}}{\text{. is given by,}}$$

               $${a_n} = a + \left( {n - 1} \right)d \ldots \left( 1 \right)$$

               $${\text{Where,}}$$ $$a = $$ $${\text{First term of A}}{\text{.P}}{\text{.,}}$$ $$d = $$ $${\text{common difference of A}}{\text{.P}}{\text{.,}}$$ $${a_n} = $$ $${{\text{n}}^{{\text{th}}}}{\text{term of an A}}{\text{.P}}{\text{.}}$$ 

               $$n = $$ $${\text{Total number of terms in A}}{\text{.P}}{\text{.}}$$

               $${\text{Substitute the known values in equation }}\left( 1 \right)$$

               $$ \Rightarrow 693 = 306 + \left( {n - 1} \right)9$$

               $$ \Rightarrow 693 - 306 = \left( {n - 1} \right)9$$

               $$ \Rightarrow 387 = \left( {n - 1} \right)9$$

               $$ \Rightarrow \left( {n - 1} \right) = \dfrac{{387}}{9}$$

               $$ \Rightarrow n - 1 = 43$$

               $$ \Rightarrow n = 44$$

               $${\text{There are total 44 numbers between 300 to 700 which are multiple of 9}}{\text{.}}$$

$${\textbf{Step 2: Find the sum of multiple of 9 lying between 300 to 700.}}$$

               $${\text{Sum of the n terms of an A}}{\text{.P}}{\text{. having first term 'a' and last term 'l' is given by, }}$$

               $${S_n} = \dfrac{n}{2}\left( {a + l} \right)$$

               $$ \Rightarrow {S_{44}} = \dfrac{{44}}{2}\left( {306 + 693} \right)$$

               $$ \Rightarrow {S_{44}} = 22\left( {999} \right)$$

               $$ \Rightarrow {S_{44}} = 21978$$

$${\textbf{Final Answer: Hence, sum of all multiple of 9 lying between 300 to 700 is 21978.}}$$

                             $${\textbf{Therefore, option (A) 21978 is correct answer.}}$$

Given series is $$3,3+\sqrt2, 3+2 \sqrt 2, 3+3 \sqrt2,.......$$

Clearly, rose plants in the rows from an AP $$23,21,19,...,5$$ with first term $$a=23$$ and common difference $$d=-2$$.
Let there be $$n$$ rows. Then,
$$a_n=5\Rightarrow a+(n-1)d$$

Sum of the term $$ S_{n} = \dfrac{n}{2}\left[2\times a + \left(n-1 \right)d \right]$$
Given sum of $$9$$ terms $$= 81$$
$$ 81 = \dfrac{9}{2} \left[2a + \left(9-1 \right)d \right] ...(1)$$
Sum of $$20$$ terms $$= 400$$
$$ 400 = \dfrac{20}{2}\left[2a+\left(20-1 \right)d \right]...(2)$$
$$\Rightarrow 81 = \dfrac{9}{2} \left[2a+8d\right]$$
$$\Rightarrow 400 = \dfrac{9}{2} \left[2a+19d\right]$$
$$\Rightarrow 162 = 18a+72d...(3)$$
$$\Rightarrow 400=20a+190d....(4)$$
Multiply eq$$(3)$$ by $$20$$ and and eq$$(4)$$ by $$18$$ and subtract both the equation
$$\Rightarrow 360a+3420d = 7200$$
$$\Rightarrow 360a+1440d = 3240$$
$$1980d = 3690$$
$$\Rightarrow d = 2$$
Now substitute the $$d = 2$$ in eq$$(4)$$
$$\Rightarrow 400= 20a+190\times 2$$
$$\Rightarrow 400 = 20a+380$$
$$\Rightarrow 20a = 20$$
$$\Rightarrow a =1$$
Sum of the 15th Term
$$S_{15} = \dfrac{15}{2} \left[2\times 1+\left(15-1 \right)2 \right]$$
$$S_{15} = \dfrac{15}{2} \left[2+\left(14\right)2 \right]$$
$$S_{15} = \dfrac{15}{2}\left[2+28 \right]$$
$$S_{15} = \dfrac{15}{2} \left[30\right]$$
$$S_{15} = 15 \times 15 = 225$$
Hence $$a =1 , d = 2 , S_{15} = 225$$

$$S_{n}=16500, n=10, d=100$$
$$S_{n}=\dfrac{n}{2}[2a+(n-1)d]$$
$$\therefore 16500=\dfrac{10}{2}[2a+(10-1)100]$$
$$16500=5[2a+900]$$
$$16500=10a+4500$$
$$10a=12000$$
$$\therefore a=1200$$
Hence he saves Rs. $$1200$$ in first year.

Given, $$a=-14$$ and $$a_5=2$$.
$$\Rightarrow a+4d=2\Rightarrow 4d=2+14\Rightarrow d=4$$
 Let the given sequence has $$n$$ terms. Therefore,
 $$S_n=\dfrac n2[2a+(n-1)d]=40$$           (given)
$$\Rightarrow \dfrac n2[2\times-14+(n-1)4]=40$$ 
$$\Rightarrow n[-14+(n-1)2]=40$$ 
$$\Rightarrow n(2n-16)=40$$ 
$$\Rightarrow n^2-8n-20=0$$ 
$$\Rightarrow (n-10)(n+2)=0$$
$$\Rightarrow n=10,-2$$
But, $$n=-2$$ is not possible.
Therefore $$n=10$$.

Here the first term is $$a$$ and the common difference is $$d$$.
$$n^{th}$$ term of the AP is given by,

Given, $$a =5, d =3, a_n = 50$$
$$\Rightarrow a + (n -1)d = 50$$
$$\Rightarrow 5 + (n -1)3 = 50$$
$$\Rightarrow 5 + 3n -3 = 50\Rightarrow 3n = 48\Rightarrow n =16$$
$$\therefore S_{16} = \dfrac{16}{2} [2a + (16-1)d]=8[2\times5+15\times3]=440$$
Hence, $$n=16,S_{16}=440$$