Question

find the middle terms of sequence formed by all numbers 9 and 95 which leaves a remainder 1 when divided by 3 . Also , find sum of nubers on both sides of middle term seperately

Answer 1

starting number leaving remainder 1 when divided by 3= 10
last number = 94
so the sequence is
10,13,16,19,22------------------------94
this becomes an AP
here starting no. a=10
last no.
$a {}^{n} = 94$
d=3
formula is
$a {}^{n} = a + (n - 1) \times d$
94=10+(n-1)3
94-10=(n-1)3
84=(n-1)3
84/3=n-1
28+1=n
n=29
middle term is
$(n + 1) \frac{1}{2} {}^{th} term$
(29+1)/2=30/2=15
middle term is 15th term
$a {}^{15 }= 10 + (15 - 1)3 \\ \: \: \: \: \: \: \: = 10 + 14 \times 3 \\ \: \: \: \: \: \: \: = 10 + 42 \\ \: \: \: \: \: \: \: = 52$

Here sum of first 14 terms are
a=10 , l=49, n=14
sum formula is
$s {}^{n} = \frac{n}{2} (2a + (n - 1)d) \\ \: \: \: \: \: \: = \frac{14}{2} (2 \times 10 + (14 - 1)3) \\ \: \: \: \: \: \: = 7(20 + 39) \ \\ \: \: \: \: \: \: = 7 \times 59 \\ \: \: \: \: \: \: = 413$
other part you solve yourself