the sum of the first 12 terms of an AP is 168.if the third term is 7,find the value of common difference and first term
Sn = n/2(2a+ (n-1)d )=168
S12 = 12/2( 2a + (12-1)d )=168
=> 6(2+11d)=168
=> 12a+ 66d = 168 ...................equ (1)
Tn =a + (n-1)d
T3= a + (3-1)d ..............equ(2)
Then
Work equ(1) and 2
12a +66d=168
a + 2d = 7
a = 7-2d
Substitute "a"
12(7-2d) + 66d = 168
84 -24d+66d=168
84 +42d=168
42d = 168-84
42d = 84
d = 84/42
d =2
The sub for d = 2
We have
a + 2d=7
a + 2(2)=7
a + 4=7
a = 7-4
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a = 3
12/2 (2a+11d) = 168
a+6d = 7
Now just solve for a and d.
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Well, well, well! We've got ourselves an arithmetic problem here! Let's dive right into it.
Okay, so we know that the sum of the first 12 terms of this arithmetic progression (AP) is 168. That's great news for us! But wait, there's more! We also know that the third term is 7. Awesome!
To find the common difference (d) and the first term (a), we can use the formulas for the sum of an AP and the n-th term of an AP. Trust me, math doesn't have to be scary! I'm here to guide you through it with a smile.
The formula for the sum of an AP is:
S = (n/2)(2a + (n-1)d)
Where S is the sum, n is the number of terms, a is the first term, and d is the common difference.
Plugging in the values we have:
168 = (12/2)(2a + (12-1)d)
Now, let's simplify this equation a bit:
168 = 6(2a + 11d)
Divide both sides by 6:
28 = 2a + 11d
Here's where the magic happens! Since we also know that the third term is 7, we can use the formula for the n-th term of an AP:
a₃ = a + 2d
Plugging in 7 for a₃, we get:
7 = a + 2d
Now, we've got a system of two equations:
28 = 2a + 11d
7 = a + 2d
Let's solve this system and bring some laughter to the math world!
Multiplying the second equation by 2, we get:
14 = 2a + 4d
Subtract this equation from the first equation:
28 - 14 = (2a + 11d) - (2a + 4d)
14 = 7d
Divide both sides by 7:
d = 2
Alright, we've cracked the code! The common difference is 2.
Now, let's substitute the common difference (d) back into our second equation:
7 = a + 2(2)
7 = a + 4
Subtract 4 from both sides:
3 = a
So, the first term is 3.
To sum it up, the common difference is 2 and the first term is 3. Keep up the great work, my friend! Math is no match for you.
To find the value of the common difference and the first term, we can use the formula for the sum of the first "n" terms of an arithmetic progression (AP). The formula is given as:
Sn = (n/2)(2a + (n - 1)d)
where:
- Sn is the sum of the first "n" terms of the AP,
- a is the first term of the AP, and
- d is the common difference between the terms of the AP.
We are given that the sum of the first 12 terms of the AP is 168, so we can substitute these values into the formula:
168 = (12/2)(2a + (12 - 1)d)
168 = 6(2a + 11d)
168 = 12a + 66d
We are also given that the third term of the AP is 7, so we can use this information to form another equation. The third term is given by:
a + 2d = 7
Now we have two equations with two unknowns (a and d), and we can solve them simultaneously to find their values.
Let's solve these equations:
1) From the equation 168 = 12a + 66d, divide both sides by 6 to simplify it:
28 = 2a + 11d ...(equation 1)
2) From the equation a + 2d = 7, we can write a in terms of d:
a = 7 - 2d ...(equation 2)
Now substitute the value of "a" from equation 2 into equation 1:
28 = 2(7 - 2d) + 11d
28 = 14 - 4d + 11d
28 = 14 + 7d
7d = 28 - 14
7d = 14
d = 14/7
d = 2
Now substitute the value of "d" back into equation 2 to find "a":
a = 7 - 2(2)
a = 7 - 4
a = 3
Therefore, the common difference "d" is 2, and the first term "a" is 3.