Question

please answer
I will mark as brainlist ​

Answer 1

$\large\bf\blue{Solution\:of\:5th➠}$

$\large\bf\pink{\bigstar Given➠}$

Sum of 10th and 11th term of A.P is 65.

means,

10th term+11th term=65

(a+9d)+(a+10d)=65

2a+19d=65

$\large\bf\blue{Solution\:a)➛}$

(Sum of 1st and 20th term of A.p)

1st term=a

20th term=a+19d.

Sum=a+a+19d

Sum=2a + 19d

(value of 2a + 19d is 65 given above)

hence,

Sum of 1st and 19th term = 65

$65 \: \: \: \: \: \: \: \: \red \bigstar \: answer\red \bigstar$

_______________________________

$\large\bf\blue{Solution\:b)➛}$

(sum of first 20th term of A.p)

1st term= a

last term(20th term)=a+19d

formula to be used(when last term is given)➠

$sum \: of \: series = \frac{n}{2} \times (a + t)$

$sum \: of \: series = \frac{n}{2} \times (a + a + 19d)$

$sum \: of \: series = \frac{n}{2} \times (2a + 19d)$

$sum \: of \: series = \frac{20}{2} \times (2a +19d)$

$sum \: of \: series = 10\times (65)$

$650 \: \: \: \: \red \bigstar \: answer \: \red \bigstar$

___________________________________

$\large\bf\blue{Solution\:c)➛}$

(17th term of A.P)

a+3d=13................(4th term)

2a + 19d=65

Comparing both,

we get,

a= 1

d=4

Now,

17th term=a+16d

=1 +(16×4)

=1+64

=65....

$\red \bigstar \: answer \: \red \bigstar$

___________________________________

$\large\bf\blue{Solution\:d)➛}$

(common difference)

Allredy Solved in Question 3rd..

$d = 4\: \: \: \red \bigstar \: answer \: \red \bigstar$