Question

an arithmetic progression consists of 37 terms. The sum of the first 3 terms of ot is 12 and the sum of the last 3 terms is 318, then find the first and last terms of the progression ​

Answer 1

Answer:

• First term of the progression is 1.
• Last term of the progression is 109.

Step-by-step explanation:

Given that:

• An arithmetic progression consists of 37 terms.
• The sum of the first 3 terms of is 12.
• The sum of the last 3 terms is 318.

To Find:

• The first and last terms of the progression.

Formula used:

• aₙ = a + (n - 1)d

Where,

• aₙ = nth terms
• a = First term
• d = Common difference

Let us assume:

• First term be a.
• Common difference be d.
• Last term be l.

Sum of the first 3 terms:

⟿ a + a + d + a + 2d = 12

⟿ 3(a + d) = 12

⟿ a + d = 12/3

⟿ a + d = 4

⟿ a = 4 - d ______(i)

Sum of the last 3 terms:

⟿ l + l - d + l - 2d = 318

⟿ 3(l - d) = 318

⟿ l - d = 318/3

⟿ l - d = 106

Last term is 37.

⟿ a₃₇ - d = 106

⟿ a + 36d - d = 106

⟿ a + 35d = 106

Substituting the value of a from eqⁿ(i).

⟿ 4 - d + 35d = 106

⟿ 34d = 106 - 4

⟿ 34d = 102

⟿ d = 102/34

⟿ d = 3 ______(ii)

Finding the first term:

Now in equation (i).

⟿ a = 4 - d

Substituting the value of d from eqⁿ(ii).

⟿ a = 4 - 3

⟿ a = 1

∴ First term = 1

Finding the last term:

⟿ l = a₃₇

⟿ l = a + (37 - 1)d

Substituting the value of a and d.

⟿ l = 1 + (36 × 3)

⟿ l = 1 + 108

⟿ l = 109

∴ Last term = 109

Answer 2

Given:-

• No. of terms ( n ) = 37

• Sum of first three terms = 12

• Sum of last three terms = 318

To Find:-

• First and last term of the progression

Solution:-

Firstly,

➾ Sum of first three terms = 12

➾ a + a+d + a+2d = 12

➾ 3a + 3d = 12

➾ 3(a + d) = 12

➾ a + d = $\dfrac{12}{3}$

➾ a + d = 4

➾ a = 4 - d ____ { 1 }

Now,

➾ Sum of last three terms = 318

➾ $a_{37} + a_{36} + a_{35}$ = 318

➾ a+36d + a+35d + a+34d = 318

➾ 3a + 105d = 318

➾ 3(a + 35d) = 318

➾ a + 35d = 106

Putting value of "a" from {1} in it,

➾ 4-d + 35d = 106

➾ 34d = 102

➾ d = $\dfrac{102}{34}$

➾ d = 3

Now, putting the value of "d" in equ. {1},

➾ a = 4 - d

➾ a = 4 - 3

➾ a = 1

Hence, the first term of A.P is 1.

➾ last term of an A.P = a + 36d

➾ $a_{37}$ = 1 + 36×3

➾ $a_{37}$ = 1 + 108

➾ ${\bold{a_{37}}}$ = 109

Hence, the last term of an A.P is 109.

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