Arithmetic progression class 10 word problems

Given $a_n = 5 - 6n$
$a_1= 5-6=-1$
$a_2= 5-12=-7$
Common difference = $a_2 - a_1= -7 +1 = -6$
Now
$S_m= \frac {m}{2}[2a + (m-1)d]$
$S_m= \frac {m}{2} [-2 + (m-1)-6]= m (2 - 3m)$

The series will be 102, 104,106,....998
Now a=102, d=2 ,n=?
Now from nth term formula
$a_n= a + (n-1)d$
$998 = 102 + (n-1)2$
$n=449$
Now
$S_n= \frac {n}{2}[2a + (n-1)d]$
$=\frac {449}{2}[204 + (449-1)2]=246950$

Given
$ a_5 = 30$, $a_12 = 65$
Now
$ a_5 = 30$
$ a + (5 - 1)d=30$
$a+ 4d=30$ -(1)


Also $a_12 = 65$
$ a + (12 - 1)d = 65 $
$a + 11d = 65$ -(2) Solving (1) and (2)
a=10, d=5

Sum of nth terms
$S_n = \frac {n}{2} [2a + (n � 1) d]$
$ S_{20} = \frac {20}{2} [2(a) + (20 � 1) d] $

$S_{20} = 1150 $

Sum of nth terms
$S_n = \frac {n}{2} [2a + (n � 1) d]$

$S_7 = \frac {7}{2} [2a + 6d]=49$
or
$2a + 6d=14$--(1)

$S_{17} = \frac {17}{2} [2a + 16d]=289$
or
$2a + 16d=34$ ---(2)
Solving (1) and (2)
a=1 ,d=2
Now
Sum of nth terms
$S_n = \frac {n}{2} [2a + (n � 1) d]$

$=\frac {n}{2} [2 + (n � 1) 2]=n^2$

$T_{25} = S_{25} - S_{24}$ $=\frac {3(25)^2}{2} + \frac {13 \times 25}{2} - \frac {3(24)^2}{2} + \frac {13 \times 24}{2}=80$

Sum of nth terms
$S_n = \frac {n}{2} [2a + (n � 1) d]$
40 installment =3600
$S_{40} = \frac {40}{2} [2a + 39d]=3600$
or
$2a + 39d=180$--(1)

30 installment = 3600 - (1/3) x 3600 = 2400
$S_{30} = \frac {30}{2} [2a + 29d]=2400$
or
$2a + 29d=160$ ---(2)
Solving (1) and (2)
a=51 ,d=2

For first tree =20 m distance is covered in to fro from well
For second tree =$2(10+ 1\times 5)=30$ m distance is covered in to fro from well
For third tree =$2(10+2 \times 5)=40$ m distance is covered in to fro from well
....
For 25th tree= $2(10 + 24 \times 5)=260$ m distance is covered in to-fro from well
So total distance
=20 + 30 + 40 +....+ 260
This is a A.P with a=20,d=10 and n=25
$S_n = \frac {n}{2} [2a + (n � 1) d]$
$S= \frac {25}{2}[40 + 24 \times 10]=3500$ m

Amount counted in half an hour = Rs 180 * 30 = Rs.5400
Therfore the remaining to be calculated is = Rs10710 - Rs 5400 = Rs 5310
According to the question, after 30 min, counting rate start decreasing every minute by 3
So At 31st min ,counting rate is 177

At 32nd min, counting rate will be 174
At 33nd min, counting rate will be 171

This is clearly as A.P
177, 174,171.......
Here a=177, d=-3 ,n=?,$S_n=5310$
$S_n = \frac {n}{2} [2a + (n � 1) d]$
$5310=\frac {n}{2} [354 + (n-1)-3]$
$ 3n^2 -357n +10620 = 0$
By using quadratic formula
$n=\frac {-b \pm \sqrt {b^2 -4ac}}{2a}$
n= 59 or 60. so we will use 59 because these are minutes and itis the work is in 59 min.

So total time to count all the Rs= 59min + 30min = 1hr 29min

Let first prize is a,then seven prizes will be
a, a-20,a- 40,a-60,a-80,a-100,a-120

Now
$a+ a-20+a- 40+a-60+a-80+a-100+a-120=700$
$7a=1120$
$a= 160$
So prizes will be
160,140,120,100,80,60,40

a=8
$a_n=33$
$S_n=123$
Now
$S_n = \frac {n}{2} [2a + (n � 1) d]$

$S_n=\frac {n}{2} [a + a_n]$
$123= \frac {n}{2} [8 + 33]$
n=6
Also
$a_n= a + (n-1)d$
$33=8+ (6-1)d$
d=5

a=17
$a_n=350$
d=9
$a_n= a + (n-1)d$
$350 = 17 + (n-1)9$
n=38
Now
$S_n=\frac {n}{2} [a + a_n]$
$=\frac {38}{2} [17 + 350]=6973$

a=2
$a_n=29$
$S_n=155$
Now
$S_n=\frac {n}{2} [a + a_n]$
$155= \frac {n}{2} [2+ 29]$
n=10

Now
$a_n= a + (n-1)d$
$29=2 + (10-1)d$
d=3

a = 6, d = -4

Let the four consecutive numbers in AP be (a - 3d), (a - d), (a + d) and (a + 3d) According to the question.
$a-3d + a - d + a + d + a + 3d = 32 $
$4a = 32$
$a = 8$ ......(1)
Now,
$\frac {(a - 3d)(a + 3d)}{(a - d)(a + d)} = \frac {7}{15}$
$15(a^2 - 9d^2) = 7(a^2 - d^2) $
$15a^2 - 135d^2 = 7a^2 - 7d^2$
$8a^2 = 128d^2$
Putting the value of a = 8 in above we get.
$ 8(8)^2 = 128d^2$
$d^2 = 4$
$d=\pm 2$

Putting a=8, d=+2
the four consecutive numbers are 2, 6, 10 and 14
Putting a=8 ,d=-2 ,we get
the four consecutive numbers are 14, 10, 6 and 2