Question

the sum of an AP 11, 17, 23 .... upto n terms is 700. how many terms are there in the series

Answer 1

Given:-

• An A.P = 11, 17, 23, . . . .

• Sum of the given A.P upto n terms = 700.

To Find:-

• No. of terms in the given A.P.

Formula Used:-

• ${\boxed{\bf{S_n = \dfrac{n}{2}[2a+(n-1)d]}}}$

Here,

• $\bf S_n$ = Sum of n terms

• a = First term of the A.P

• d = Common Difference

• n = No. of terms in an A.P

Solution:-

Using Formula,

$\bf :\implies\:S_n = \dfrac{n}{2}[2a+(n-1)d]$

Here,

• $\bf S_n=700$

• a = 11

• d = 17 - 11 = 6

Putting values,

$\sf :\implies\:700 = \dfrac{n}{2}[2\times11+(n-1)6]$

$\sf :\implies\:700\times2= n[22+6n-6]$

$\sf :\implies\:1400= n[16+6n]$

$\sf :\implies\:6n^2+16n= 1400$

$\sf :\implies\:6n^2+16n-1400= 0$

$\sf :\implies\:3n^2+8n-700= 0$

Now, Using Middle Term Split:-

$\sf :\implies\:3n^2+(50-42)n-700=0$

$\sf :\implies\:3n^2+50n-42n-700=0$

$\sf :\implies\:n(3n+50)-14(3n+50)=0$

$\sf :\implies\:(3n+50)(n-14) =0$

$\sf :\implies\:3n+50 = 0\:and\:n-14 =0$

$\sf :\implies\:n = \dfrac{50}{3}\:and\:n=14$

{n cannot be in fraction} So,

$\bf :\implies\:n=14$

Hence, There are 14 terms in the series whose sum is 700.

━━━━━━━━━━━━━━━━━━━━━━━━━

Answer 2

.To Find:-

No. of terms in the given A.P.

Formula Used:-

{\boxed{\bf{S_n = \dfrac{n}{2}[2a+(n-1)d]}}}

S

n

=

2

n

[2a+(n−1)d]

Here,

\bf S_nS

n

= Sum of n terms

a = First term of the A.P

d = Common Difference

n = No. of terms in an A.P

Solution:-

Using Formula,

\bf :\implies\:S_n = \dfrac{n}{2}[2a+(n-1)d]:⟹S

n

=

2

n

[2a+(n−1)d]

Here,

\bf S_n=700S

n

=700

a = 11

d = 17 - 11 = 6

Putting values,

\sf :\implies\:700 = \dfrac{n}{2}[2\times11+(n-1)6]:⟹700=

2

n

[2×11+(n−1)6]

\sf :\implies\:700\times2= n[22+6n-6]:⟹700×2=n[22+6n−6]

\sf :\implies\:1400= n[16+6n]:⟹1400=n[16+6n]

\sf :\implies\:6n^2+16n= 1400:⟹6n

2

+16n=1400

\sf :\implies\:6n^2+16n-1400= 0:⟹6n

2

+16n−1400=0

\sf :\implies\:3n^2+8n-700= 0:⟹3n

2

+8n−700=0

Now, Using Middle Term Split:-

\sf :\implies\:3n^2+(50-42)n-700=0:⟹3n

2

+(50−42)n−700=0

\sf :\implies\:3n^2+50n-42n-700=0:⟹3n

2

+50n−42n−700=0

\sf :\implies\:n(3n+50)-14(3n+50)=0:⟹n(3n+50)−14(3n+50)=0

\sf :\implies\:(3n+50)(n-14) =0:⟹(3n+50)(n−14)=0

\sf :\implies\:3n+50 = 0\:and\:n-14 =0:⟹3n+50=0andn−14=0

\sf :\implies\:n = \dfrac{50}{3}\:and\:n=14:⟹n=

3

50

andn=14

{n cannot be in fraction} So,

\bf :\implies\:n=14:⟹n=14

Hence, There are 14 terms in the series whose sum is 700.