Question

Jinal saves Rs 1600 during the first year , Rs 2100 in the second year , Rs 2600 in the third year. If she continues her saving in this pattern , in how many years will she save Rs 38500 ?​

Answer 1

Answer:

10 years.

Step-by-step explanation:

Jinal's savings:

First year =Rs.1600

Second year =Rs.2100

Third year =Rs.. 2600

Sequence : 1600,2100,2600 ...

So, a = first term = 1600

Common difference = d = 2100-1600=2600-2100=500

$So,sum of first n terms = S_n=\frac{n}{2}(2a+(n-1)d)Sn=2n(2a+(n−1)d)</p><p>So, to find . in how many years will she save rs,. 38500</p><p>38500=\frac{n}{2}(2(1600)+(n-1)500)38500=2n(2(1600)+(n−1)500)</p><p>77000=2700n+500n^277000=2700n+500n2</p><p>n=\frac{-77}{5} ,10n=5−77,10$

Hence she will save Rs.38500 in 10 years.

Answer 2

Step-by-step explanation:

Answer:

10 years.

Step-by-step explanation:

Jinal's savings:

First year =Rs.1600

Second year =Rs.2100

Third year =Rs.. 2600

Sequence : 1600,2100,2600 ...

So, a = first term = 1600

Common difference = d = 2100-1600=2600-2100=500

So,sum of first n terms = S_n=\frac{n}{2}(2a+(n-1)d)Sn=2n(2a+(n−1)d) < /p > < p > So, to find . in how many years will she save rs,. 38500 < /p > < p > 38500=\frac{n}{2}(2(1600)+(n-1)500)38500=2n(2(1600)+(n−1)500) < /p > < p > 77000=2700n+500n^277000=2700n+500n2 < /p > < p > n=\frac{-77}{5} ,10n=5−77,10So,sumoffirstnterms= S

n

=

2

n

$Answer:</p><p></p><p>10 years.</p><p></p><p>Step-by-step explanation:</p><p></p><p>Jinal's savings:</p><p></p><p>First year =Rs.1600</p><p></p><p>Second year =Rs.2100</p><p></p><p>Third year =Rs.. 2600</p><p></p><p>Sequence : 1600,2100,2600 ...</p><p></p><p>So, a = first term = 1600</p><p></p><p>Common difference = d = 2100-1600=2600-2100=500</p><p></p><p>So,sum of first n terms = S_n=\frac{n}{2}(2a+(n-1)d)Sn=2n(2a+(n−1)d) < /p > < p > So, to find . in how many years will she save rs,. 38500 < /p > < p > 38500=\frac{n}{2}(2(1600)+(n-1)500)38500=2n(2(1600)+(n−1)500) < /p > < p > 77000=2700n+500n^277000=2700n+500n2 < /p > < p > n=\frac{-77}{5} ,10n=5−77,10So,sumoffirstnterms= Sn=2n(2a+(n−1)d)Sn=2n(2a+(n−1)d)</p><p>So,tofind.inhowmanyyearswillshesavers,.38500</p><p>38500=2n(2(1600)+(n−1)500)38500=2n(2(1600)+(n−1)500)</p><p>77000=2700n+500n277000=2700n+500n2</p><p>n=5−77,10n=5−77,10</p><p></p><p>Hence she will save Rs.38500 in 10 years</p><p></p><p>$

(2a+(n−1)d)Sn=2n(2a+(n−1)d)</p><p>So,tofind.inhowmanyyearswillshesavers,.38500</p><p>38500=

2

n

(2(1600)+(n−1)500)38500=2n(2(1600)+(n−1)500)</p><p>77000=2700n+500n

2

77000=2700n+500n2</p><p>n=

5

−77

,10n=5−77,10

Hence she will save Rs.38500 in 10 years