Question

15th term and 11th term of an A.P are 3 and 23 respectively .Is 40 a term of the sequence .

Answer 1

Here,
$a + (15 - 1)d = 3 \\ a + (11 - 1)d = 23$

$(a + 14d) =( a + 10d) - 20 \\ 4d = (- 20) \\ d = ( - 5)$

Also, from above, A=73
Now,

$40 = 73 + (n - 1)( - 5) \\ n = \frac{38}{5}$

Since, no Integral value of N, so 40 is not a term of the A.P.

Hope It Helps!!

Answer 2

Answer:

No

Step-by-step explanation:

Here,

\begin{gathered}a + (15 - 1)d = 3 \\ a + (11 - 1)d = 23\end{gathered}a+(15−1)d=3a+(11−1)d=23

(a+14d)=(a+10d)−204d=(−20)d=(−5)Also, from above, A=73

Now,

\begin{gathered}40 = 73 + (n - 1)( -

5) \\ n = \frac{38}{5} \end{gathered}40=73+(n−1)(−5)n=538