A man repays a debt of Rs 4860 by paying Rs 30 in the first month and then increases the amount by Rs 15 every month.How long will it take time to clear the debt(Assume that no interest is charged)?
Answers
Answer:
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Step-by-step explanation:
Arithmetic sequences are the sequences which have a constant difference between the numbers. In other words, it is the sequence of numbers in which we obtain the next term by adding a constant number.So, it is in the form of
(
a
,
a
+
d
,
a
+
2
d
,
a
+
3
d
...
.
For example,
↦
3
,
5
,
7
,
9
,
11
,
13
...
Is an arithmetic sequence. In this the common difference is
2
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Now, let's understand this question. In this, he pays
3250
by giving
20
in the first month. And then, increases it by
15
every month thereafter.
↦
20
,
35
,
50
,
65
...
.
In this manner he clears the loan in
n
months
To solve this, we need to know the formula
S
n
=
n
2
(
2
a
+
(
n
−
1
)
d
)
Where,
S
n
is the sum of
n
number of terms.
a
is the first term(
20
). And,
d
is the common difference (In this case, it is
15
). Finally,
n
is the number of terms
According to the question, We need to find
n
. And also, they have given
S
n
=
3250
So,
→
S
n
=
n
2
(
2
a
+
(
n
−
1
)
d
)
→
3250
=
n
2
(
2
(
20
)
+
(
n
−
1
)
(
15
)
)
→
n
(
40
+
15
n
−
15
)
=
2
×
3250
→
n
(
25
+
15
n
)
=
6500
→
15
n
2
+
25
n
−
6500
=
0
→
3
n
2
+
5
n
−
1300
=
0
→
(
n
−
20
)
(
3
n
+
65
)
=
0
→
n
−
20
=
0
⇒
n
=
20
months
Answer:
Suppose the loan is cleared in n months. Clearly, the amounts form an A.P. with first term 20 and the common difference 15.
∴ Sum of the amounts =3250
⟹
2
n
{2×20+(n−1)×15}=3250
⟹
2
n
(40+15n−15)=3250
⟹n(15n+25)=6500
⟹15
2
+25n−6500=0
⟹3n
2
+5n−1300=0
⟹(n−20)(3n+65)=0
⟹n=20 or, n=−
3
65
⟹n=20 [∵n
=
3
65
]
Thus, the loan is cleared in 20 months