Question

I NU UIL Ju term of an arithmetic progression 2, 6, 10, 14, .
5. al
18. Find the sum of first 20 terms of the arithmetic progression 3, 8, 13,
using the formula.
OR​

Answer 1

Answer:

a=2

d=6-2=4

n=20

s=n/2[2a+(n-1).d]

s=20/2[2.2+(20-1).4]

s=10[4+(19).4]

s=10[4+76]

s=10[80]

s=10*80

s=800

is my ans is correct or not

Step-by-step explanation:

Answer 2

Answer :

Step by Step Explanation :

${10}^{th} \: \: term \: \: of \: \: AP : 2, \: 6, \: 10, \: 14 \\ Here \: \: a = 2, \: \: d = 6 - 2 = 4 \: \: n = 10 \\ T_{n} = a + (n - 1)d \\ T_{10} = 2 + (10 - 1)4 \\ T_{10} = 2 + (9)4 \\ T_{10} = 2 + 36 \\ \boxed{ T_{10} = \underline{ \underline{ 38}}} \\ \\ Sum \: \: of \: \: first \: \: 20 \: \: terms \: \: of \: \: AP : 3, \: 8, \: 13 \\ Here \: \: a = 3, \: \: d = 8 - 3 = 5 \: \: n = 20 \\ S_{n} = \frac{n}{2} [2a + (n - 1)d] \\ S_{20} = \frac{20}{2} [2 \times 3 + (20 - 1)5] \\ S_{20} = 10 [6 + (19)5] \\ S_{20} = 10 [6 + 95] \\ S_{20} = 10 [101] \\ \boxed{ S_{20} = \underline{ \underline{ 1010}}}$

[]