Find the number of term in the Arithmetic Progression(A.P) 2, -9, -20,...-141.
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Correct Answer: Option D
Explanation:
T1, T2, T3
2, -9, -20 .... -141
l = a + (n - 1)d
first term, a = 2
common difference d = T3 - T2
= T2 - T1
= -20 - (-9) = -9 -2
= -20 + 9
= -9 -2
= -20 + 9
= -11
-11 = -11
d = -1
last term l = -141
-141 = 2 + (\(\cap\) - 1) (-11)
-141 = 2 + (-11 \(\cap\) + 11)
= 2 - 11\(\cap\) + 11
-141 = 13 - 11\(\cap\)
-141 - 13 = -11\(\cap\)
-154 = -11\(\cap\)
\(\cap\) = \(\frac{-154}{-11}\)
\(\cap\) = 14
T1, T2, T3
2, -9, -20 .... -141
l = a + (n - 1)d
first term, a = 2
common difference d = T3 - T2
= T2 - T1
= -20 - (-9) = -9 -2
= -20 + 9
= -9 -2
= -20 + 9
= -11
-11 = -11
d = -1
last term l = -141
-141 = 2 + (\(\cap\) - 1) (-11)
-141 = 2 + (-11 \(\cap\) + 11)
= 2 - 11\(\cap\) + 11
-141 = 13 - 11\(\cap\)
-141 - 13 = -11\(\cap\)
-154 = -11\(\cap\)
\(\cap\) = \(\frac{-154}{-11}\)
\(\cap\) = 14