Question

For an arithmetic squence,9th term is 16 and it's 16th term is 9. (a)what is the common difference? (b)what is the 25th term? (c)what is the sum of first 49 terms?​

Answer 1

$\tt{1. \: \: Common \: difference = 0}$

$\tt{2. \: \: 25th \: term= a}$

$\tt{3. \: \:Sum \: of \: first \: 49 \: terms= 49a}$

Step-by-step explanation:

★Given -

• In an arithmetic sequence,9th term is 16 and it's 16th term is 9.

★To find -

1. The common difference.
2. 25th term.
3. Sum of first 49 terms.

★Solution -

It is given that,

$\tt{1. \: \: \large{a {\tiny9} = a \tiny16}}$

$\longmapsto \tt{\large{a + 8d = a + 15d}}$

$\longmapsto \tt{\large{0 = \cancel{a} + 15d \cancel{- a} - 8d}}$

$\longmapsto \tt{\large{0 = 7d}}$

$\longmapsto \tt{ \blue{\boxed{ \green{\large{\tt{0 = d}}}}}}$

$\tt{2. \: \:\large{ a {\tiny25} = a + 24d}}$

$\longmapsto \tt{\large{a {\tiny25} = a + 24(0)}}$

$\longmapsto \tt{ \red{\boxed{ \pink{ \tt{\large{a {\tiny25} = a }}}}}}$

$\tt{ \large{Sum \: of \: n \: terms \: = \: \dfrac{n}{2}[2a + (n - 1)d] }}$

$\tt{3. \: \: {\Large{S} }{\tiny49} = \dfrac{49}{2}[\large{2a + (49 - 1)0]}}$

$\tt{ \longmapsto { \Large{S}} {\tiny49} = \dfrac{49}{ \cancel2}( \cancel2a )}$

$\tt{ \longmapsto {\Large{S}} {\tiny49} = \large49( a )}$

$\tt{ \longmapsto \purple{\boxed{\orange{{ \tt\Large{S} }{\tiny49} = \large 49 a }}}}$

Answer 2

Answer:
a.-1
b.1
c.-23
d.49

Step-by-step explanation: