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Zorluk: OrtaArithmetic and Geometric Progressions (AP and GP)

The 3rd3^{\text{rd}} term of an arithmetic progression (AP) is 1010 and the 7th7^{\text{th}} term is 2222. What is the sum of the first 1212 terms of the progression?

  1. A
    228228
  2. 246246Cevap
  3. C
    264264
  4. D
    444444

Cevap

The sum of the first 1212 terms of the arithmetic progression is 246246.
Using the nthn^{\text{th}} term formula Tn=a+(n1)dT_n = a + (n-1)d, the equations a+2d=10a + 2d = 10 and a+6d=22a + 6d = 22 yield d=3d = 3 and a=4a = 4. Substituting these values into S12=122[2(4)+11(3)]S_{12} = \frac{12}{2}[2(4) + 11(3)] gives 6×41=2466 \times 41 = 246.

Adım Adım Çözüm

1
Set up simultaneous equations using the nthn^{\text{th}} term formula Tn=a+(n1)dT_n = a + (n-1)d.
a+2d=10a + 2d = 10 and a+6d=22a + 6d = 22.
The 3rd3^{\text{rd}} term corresponds to n=3n=3 and the 7th7^{\text{th}} term corresponds to n=7n=7.
2
Subtract the first equation from the second to find the common difference dd.
4d=12    d=34d = 12 \implies d = 3.
Subtracting eliminates the first term aa.
3
Substitute d=3d = 3 back into the first equation to find the first term aa.
a+2(3)=10    a=4a + 2(3) = 10 \implies a = 4.
Determining the first term is necessary to calculate the sum.
4
Calculate the sum of the first 1212 terms using Sn=n2[2a+(n1)d]S_n = \frac{n}{2}[2a + (n-1)d].
S12=122[2(4)+(121)(3)]=6[8+33]=6(41)=246S_{12} = \frac{12}{2}[2(4) + (12-1)(3)] = 6[8 + 33] = 6(41) = 246.
Applying the AP sum formula with n=12n = 12, a=4a = 4, and d=3d = 3.

Anahtar Kavram

Arithmetic Progression: Finding common difference, first term, and sum of terms
Tahmini Süre:1m 30s
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