Question

Show that a1, a2, a3, ... form an A.P. where an is defined as "an"= 3 + 4n. Also find the
sum of first 15 terms.

​

Answer 1

Answer:

525.

Step-by-step explanation:

Here,

nth term is defined by 3 + 4n.

From the properties of AS, we know

• Sum of n terms from 1st term is given by n / 2 { a + l } , where a is the first term and l is the last term.

Here,

1st term = 3 + 4n

2nd term = 3 + 4( n + 1 )

3rd term = 3 + 4( n + 2 )

If the given polynomial represents an AP, twice of 2nd term must be equal to the sum of extremes.

= > 2 x 2nd term = 2[ 3 + 4( n + 1 ) ]

; sum of extremes = 3 + 3 + 4n + 4( n + 2 ) = > 3( 2 ) + 4( n + 1 )( 2 ) = > 2[ 3 + 4( n + 1 ) ]

From the above activity we found 2 times of second term is equal to sum of extremes which means that the given polynomial represents an AP.

= > First term = 3 + 4n , for n = 1

= > Last term is 3 + 4n , for n = 15

Thus,

= > Sum of 15 terms = ( 15 / 2 ) [ first term + last term ]

= > Sum of 15 terms = ( 15 / 2 ) [ { 3 + 4( 1 ) } + { 3 + 4( 15 ) } ]

= > 7.5( 6 + 4 + 4( 15 ) ]

= > 7.5 ( 70 )

= > 525

Hence the sum of first 15 terms is 525.

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Sum\:of\:15th\:term=525}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies a_{n} = 3 + 4n \\ \\\red{\underline \bold{To \: Find :}} \\ \tt: \implies show \: that \: a_{1},a_{2},a_{3} \: are \: in \: A.P \\ \\ \tt: \implies Sum \: of \: 15th \: term(s_{15}) = ?$

• According to given question :

$\tt: \implies a_{n} = 3 + 4n \\ \\ \tt{\circ \: n = 1,2,3,...} \\ \\ \bold{For \: n = 1} \\ \tt: \implies a_{1} = 3 + 4 \times 1 \\ \\ \tt: \implies a_{1} =3 + 4 \\ \\ \tt: \implies a_{1} =7 \\ \\ \bold{For \: n = 2} \\ \tt: \implies a_{2} = 3 + 4 \times 2 \\ \\ \tt: \implies a_{2} =3 + 8 \\ \\ \tt: \implies a_{2} =11 \\ \\ \bold{For \: n = 3} \\ \\ \tt: \implies a_{3} = 3 + 4 \times 3 \\ \\ \tt: \implies a_{3} =3 + 12 \\ \\ \tt: \implies a_{3} =15 \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies 2a_{2} = a_{1} + a_{3} \: \: \: \: \: (According \: to \: A.P \: property) \\ \\ \tt: \implies 2 \times 11 = 7 + 15$

$\tt: \implies 22 = 22 \\ \\ \tt \green{\therefore From \: this \: we \: can \: conclude \: that}\\ \: \: \: \: \green{\tt{a_{1},a_{2} \: and \: a_{3} \: are \: in \: A.P}} \\ \\ \bold{For \: sum \: of \: 15th \: term} \\ \green{\tt{\circ \: First \: term( a_{1} ) = 7}} \: \\ \\ \green{\circ \tt \: Common \: difference(d) = 4} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies s_{n} = \frac{n}{2} (2a + (n - 1)d) \\ \\ \tt: \implies s_{15} = \frac{15}{2} (2 \times 7 + (15 - 1) \times 4 )\\ \\ \tt: \implies s_{15} = \frac{15}{2} \times (14 + 56) \\ \\ \tt: \implies s_{15} = \frac{15}{2} \times 70 \\ \\ \green{ \tt: \implies s_{15} = 525} \\ \\ \green{\tt{\therefore Sum \: of \: 15 \: th \: term \: is \: \: 525}}$