Question

a=7, a13=35, find d and s13

Answer 1

Given that first term of Arithmetic progression

a=a1=7

thirteenth term of Arithmetic progression is

a13=35

we can use nth term formula to find the value of common difference d

$a_n=a+(n-1)d$

plug above values

$a_{13}=7+(13-1)d$

$35=7+12d$

$35-7=7-7+12d$

$28=12d$

28/12=d

7/3=d

Now we need to find sum of 13 terms so we will use sum of n terms formula

$S_n=\frac{n}{2}{2a+(n-1)d}$

plug the given values and d=7/3

$S_{13}=\frac{13}{2}{2(7)+(13-1)\frac{7}{3}}$

$S_{13}=\frac{13}{2}{14+12\frac{7}{3}}$

$S_{13}=\frac{13}{2}{14+\frac{84}{3}}$

$S_{13}=\frac{13}{2}{14+28}$

$S_{13}=\frac{13}{2}{42}$

$S_{13}=\frac{13}{1}{21}$

$S_{13}=273$

Hence final answer is $d=\frac{7}{3}$ and $S_{13}=273$

Answer 2

Answer:

d = 7/3, Sn=273

Step-by-step explanation:

First term of an AP = a = 7

Thirteenth term of an AP = 35

a + 12d = 35 ------(1)

Substitute a in eq - (1)

a + 12d = 35

(7) + 12d = 35

12d = 35 - 7

12d = 28

d = 28/12

d = 7/3

In an AP sum of the terms = n/2 ( a + an )

= 13/2 ( 7 + 35)

= 13/2 ( 42)

= 13(21)

= 273