Question

find the sum of first 24 time term of list of numbers whose and term is given by a n is equal to 3 + 2 m​

Answer 1

If an = 3+2m

Put m = 1 then

A1 =3+2 =5

Put m=2 then

A2 =3+4=7 then ap is ,

5 ,7 ,9 ,.............

First term a1 =5

Diffrence (d )=7-5 =2

Number of terms is =n = 24

Sum (Sn)= n/2 [ 2a+ (n-1)d]

Sn =24/2[2×5 + (24-1)2]

= 12(10 +46)

= 12×56

Sn =672

Answer 2

$\large\tt\pink{As,an=3+2n}$

$\large\tt\pink{So,a1=3+2=5}$

$\large\tt\pink{a2=3+2×2=7}$

$\large\tt\pink{a3=3+2×3=9}$

$\small\sf\underline\red{List\:of\:numbers\:becomes\:5,7,9,11,.....}$

$\small\sf\red{Here,7-5=9-7=11-9=2,and\:so\:on}$

$\small\sf\underline\pink{so\:it\:forms\:an\:AP\: with\:d=2}$

$\small\sf\pink{To\:find\:S24,we\:have\:n=24,a=5,d=2}$

$\therefore$$\large\tt\purple{S24=\frac{24}{2}(2×5+(24-1)×2)}$

$\longrightarrow$$\large\tt\purple{12(10+46)}$

$\longrightarrow$$\large{\boxed{\tt{\purple{672}}}}$