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# The sum of the squares of the first and fifth term of an arithmetic

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Manager
Joined: 25 Jan 2020
Posts: 59
Location: India
Concentration: Technology, Strategy
GPA: 3.2
WE: Information Technology (Retail Banking)
The sum of the squares of the first and fifth term of an arithmetic  [#permalink]

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28 May 2020, 09:32
3
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Difficulty:

25% (medium)

Question Stats:

82% (02:02) correct 18% (02:54) wrong based on 17 sessions

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The sum of the squares of the first and fifth term of an arithmetic progression is 218.If the third term is less than the first term, and the sum of these two terms is 16, find the first term?
A.10
B.11
C.12
D.13
E.14
Manager
Joined: 10 Jun 2018
Posts: 59
Re: The sum of the squares of the first and fifth term of an arithmetic  [#permalink]

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28 May 2020, 11:22
1
MBADream786 wrote:
The sum of the squares of the first and fifth term of an arithmetic progression is 218.If the third term is less than the first term, and the sum of these two terms is 16, find the first term?
A.10
B.11
C.12
D.13
E.14

Answer: D

Let a1 be first term, a3 be third term and a5 be fifth term of arithmetic progression.

$$(first term)^2$$+$$(fifth term)^2$$=218
$$(a1)^2$$+$$(a5)^2$$=218
Checking few combinations of square terms which sum upto 218, we see that 169 and 49 can be values of $$(a1)^2$$ and $$(a5)^2$$

Now a1=13 or -13 or 7 or -7 (we are not sure if $$(a1)^2$$ =169 or 49)

Also, a3<a1 and a1+a3=16
Case 1: a1=13 and a5=7,
Since a3 lies between a1 and a5 and a3<a1, this case is not possible (a3 cannot be less than a1 but greater than a5 in an arithmetic progression)

Case 2: a1=13 and a5=-7,
a3 can easily lie between a1 and a5 here. So solving a1+a3=16 with these values we get, a3=3.
Here, a1-a3=a3-a5 that is the difference between alternate terms of arithmetic progression is same. Hence this is a valid case

The other 2 cases are first term a1=7 or -7. Since both these values are not in the option, we can skip testing these.
Answer: First term a1=13
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Re: The sum of the squares of the first and fifth term of an arithmetic  [#permalink]

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28 May 2020, 11:43
1
It's quick to solve this question using options

A. 1st term=10; 3rd term= 6; 5th term = 2; sum of squares of the first and fifth term =100+4 (rejected)
B. 1st term=11; 3rd term= 5; 5th term = -1; sum of squares of the first and fifth term =121+1 (rejected)
C. 1st term=12; 3rd term= 4; 5th term = -4; sum of squares of the first and fifth term =144+16 (rejected)
D. 1st term=13; 3rd term= 3; 5th term = -7; sum of squares of the first and fifth term =169+49= 218 (Answer)

OR

$$(a+4d)^2+ a^2 = 218$$......(1)

Also, $$a+2d+a =16$$

$$a+d=8$$.......(2)

From (1) and (2)

$$(8+3d)^2+(8-d)^2 = 218$$

$$9d^2+48d+64 + d^2-16d+64 = 218$$

$$10d^2 +32d -90 = 0$$

$$5d^2+16d-45 = 0$$

$$5d^2+25d -9d -45 = 0$$

d= -5 or 9/5(Rejected)

$$a+d=8$$
$$a=8+5=13$$

MBADream786 wrote:
The sum of the squares of the first and fifth term of an arithmetic progression is 218.If the third term is less than the first term, and the sum of these two terms is 16, find the first term?
A.10
B.11
C.12
D.13
E.14
Re: The sum of the squares of the first and fifth term of an arithmetic   [#permalink] 28 May 2020, 11:43

# The sum of the squares of the first and fifth term of an arithmetic

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