Question

The sum of 45 term of an A. P. is 3195. Find its 23rd term

Answer 1

Let the first term be 'a'
Common difference be 'd'

$s45 = 3195 \\ \frac{45}{2} \times (2a + (45 - 1)d = 3195 \\ \frac{45}{2} \times (2a + 44d) = 3195 \\ 45 \times (a + 22d) = 3195 \\ a + 22d = 71 \\ a + (23 - 1)d = 71 \\ t23 = 71$

Answer 2

Step-by-step explanation:

Let the first term be 'a'

Common difference be 'd'

\begin{gathered}s45 = 3195 \\ \frac{45}{2} \times (2a + (45 - 1)d = 3195 \\ \frac{45}{2} \times (2a + 44d) = 3195 \\ 45 \times (a + 22d) = 3195 \\ a + 22d = 71 \\ a + (23 - 1)d = 71 \\ t23 = 71\end{gathered}

s45=3195

2

45

×(2a+(45−1)d=3195

2

45

×(2a+44d)=3195

45×(a+22d)=3195

a+22d=71

a+(23−1)d=71

t23=71