Question

What is a3 in an arithmetic sequence in which a10=41 and a15=61

Answer 1

$\bf \begin{array}{llll} term&value\\ -----&-----\\ a_{10}&41\\ a_{11}&41+d\\ a_{12}&(41+d)+d\\ &41+2d\\ a_{13}&(41+2d)+d\\ &41+3d\\ a_{14}&(41+3d)+d\\ &41+4d\\ a_{15}&(41+4d)+d\\ &41+5d=61 \end{array} \\\\\\ 41+5d=61\implies 5d=20\implies d=\cfrac{20}{5}\implies \boxed{d=4}\\\\ -------------------------------$

$\bf n^{th}\textit{ term of an arithmetic sequence}\\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference}\\ ----------\\ d=4\\ n=10\\ a_{10}=41 \end{cases} \\\\\\ 41=a_1+(10-1)4\implies 41=a_1+36\implies \boxed{5=a_1}$

thus

$\bf n^{th}\textit{ term of an arithmetic sequence}\\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference}\\ ----------\\ d=4\\ n=3\\ a_{1}=5 \end{cases} \\\\\\ a_3=a_1+(3-1)4\implies a_3=5+(3-1)4$

and surely you know how much that is.

Answer 2

Answer:  The required third term of the sequence is 13.

Step-by-step explanation:  We are given to find the third term in an arithmetic sequence in which the 10th term is 41 and 15th term is 61.

We know that

the nth term of an arithmetic sequence with first term a and common difference d is given by

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

According to the given information, we have

$a_{10}=41\\\\\Rightarrow a+(10-1)d=41\\\\\Rightarrow a+9d=41~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

and

$a_{15}=61\\\\\Rightarrow a+(15-1)d=61\\\\\Rightarrow a+14d=61~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

Subtracting equation (i) from equation (ii), we get

$(a+14d)-(a+9d)=61-41\\\\\Rightarrow 5d=20\\\\\Rightarrow d=\dfrac{20}{5}\\\\\Rightarrow d=4$

From equation (i), we get

$a+9\times 4=41\\\\\Rightarrow a=41-36\\\\\Rightarrow a=5.$

Therefore, the third term of the sequence is

$a_3=a+(3-1)d=5+2\times 4=5 + 8=13$

Thus, the required third term of the sequence is 13.