Check whether-150 is a term of a ap : 11,8,5,2....
Answers
Answer:
an = a + (n-1)d
-150 = 11 +(n-1)(-3)
-150 = 11 - 3n + 3
-150 = 14 - 3n
3n = 164
n = 164 / 3
n = 52.66
as n is in decimal form and therefore -150 is not a term of ap
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☆Given:
• Arithmetic progression series:
11 , 8 , 5 , 2 ....
☆ To check:
➡ whether t - 150 is a term in the given AP.
☆Formula used:
¤ tn = a + ( n - 1 ) d
where a ➡ first term
n ➡ Number of terms
d ➡ Common difference
Given is the t -150;
Lets substitute in the above formula,
- 150 = 11 + ( n - 1 ) (-3)
[ Here a is the first term and d is the second term - first term]
-150 = 11 -3n+3
-150 = 14 -3n
-3n = - 164
=> n = 164/3 = 54.6
Since The remainder Is not zero, it can't be the term of the given arithmetic series.
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