Question

Algebraic form of an arithmetic sequence is
3/4 n+ 1/4
what is the common difference of the sequence ?
what is it's 13th term ?



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Answer 1

Answer:

in the formula substitute 1 in place of n and find the 1st ter of the ap which is a=1,now find the 2nd term of the ap by substituting 2 in place of n which is a2=7/2,now find the common difference by doing a1-a2=7/2-1.therefore d=3/4

now a13=a+12d

a13=1+12(3/4)

1+36/4

40/4

=10

therefore a13=10

Step-by-step explanation:

Answer 2

Answer:

• Value of a = 1 , d = $\frac{3}{4}$
• value of $t_{13} = 10$ .  

Step-by-step explanation:

Given :- expression is $\frac{3}{4} +\frac{1}{4} *n$

To find :-

• common difference ' d ' .
• value of 13th term .

Solution :-

Step 1) Given expression is :- $\frac{3}{4} +n*\frac{1}{4}$  ---- (1)

Step 2) We know , equation for general term ,

for nth term is ,

$t_n = a+(n-1)*d$ --- (2)

and for , $t_n = \frac{3}{4} +n*\frac{1}{4}$ ---- (3)

Step 3) Comparing equation , (2) and (3) ,

we get ,

• d = $\frac{3}{4}$  ---- (4)
• a-d = $\frac{1}{4}$
• thus , a = $\frac{1}{4} + d \\\frac{1}{4} + \frac{3}{4}\\ 1$  ----- (5)

Step 4) Therefore , the value of a = 1 , d = $\frac{3}{4}$ .

Step 5) Value of 13th term ,

from equation (2) ,

$t_n = a+ (n-1)*d$

for n = 13 ,

$t_{13} = 1 + (13-1)*\frac{3}{4}\\t_{13} = 1+ (12)*\frac{3}{4}\\ t_{13} = 1+ 3*3 \\t_{13} = 10$

Summarizing the answer

• Value of a = 1 , d = $\frac{3}{4}$
• value of $t_{13} = 10$ .  

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