Question

The 2,31 and the last term of an arithmetic progression are seven over four ,a half and a negative six and a half. Find the first term and the number of terms

Answer 1

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Given, a7 = -1 and a4 = 41

☛ Let a be the first term and d be the common difference of an A.P

-1 = a + (7 – 1)d

-1 = a + 6d . . . . . . .  . . (1)

☛ And   41 = a + (4 – 1)d

41 = a + 3d . . . . . . . (2)

Now, equation (1) – equation (2)

-1 – 41 = 6d – 3d    Therefore,   d = -14

Putting d = -14 in equation (2) we get

41 = a + 3×(-14)

Therefore, a = 83

Therefore the required A.P is 83, 69, 55, 12, 41, 27, 13………

Now, from the given A.P:

a = 83,

d = 69 – 83 = -14

n = 17

a17 = ??

Since,                   an = a + (n – 1)d

Therefore          a17 = 83 + (17 – 1)× -14

a17 = 208 + 14 – 26

a17 = -141

Therefore 17th term of this given A.P is -141

Answer 2

✨Hey mate ✨

Here is ur answer ⤵️⤵️


Given, a7 = -1 and a4 = 41

Let a be the first term and d be the common difference of an A.P

-1 = a + (7 – 1)d

-1 = a + 6d . . . . . . .  . . (1)

And   41 = a + (4 – 1)d

41 = a + 3d . . . . . . . (2)

Now, equation (1) – equation (2)

-1 – 41 = 6d – 3d    Therefore,   d = -14

Putting d = -14 in equation (2) we get

41 = a + 3×(-14)

Therefore, a = 83

Therefore the required A.P is 83, 69, 55, 12, 41, 27, 13………

Now, from the given A.P:

a = 83,

d = 69 – 83 = -14

n = 17

a17 = ??


Since,                   an = a + (n – 1)d

Therefore          a17 = 83 + (17 – 1)× -14

a17 = 208 + 14 – 26

a17 = -141

Therefore 17th term of this given A.P is -141✔️



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